NEET Practice Q & A

Hint

(c) In case of \(\mathrm{AC}, Z=\sqrt{R^2+(\omega L)^2}\), while in case of DC , it is only \(R\). So, in AC resistance increases.

Hint

(c) At frequencies higher than resonant frequency, \(X_L>X_C\) i.e. circuit behaves as inductive circuit.

Hint

(c) The correct answer is decrease sinusoidally. In an RLC circuit with negligible resistance, the charge on the capacitor oscillates and decreases sinusoidally as energy transfers between the capacitor and the inductor.

Explanation:
Initial State:
The capacitor is initially charged, meaning it stores electrical energy.
Energy Transfer:
When the charging battery is disconnected, the capacitor begins to discharge, transferring its stored electrical energy to the inductor. The inductor, in turn, stores this energy as a magnetic field.

No Resistance:
In a circuit with negligible resistance, there is no mechanism for energy dissipation. Therefore, the oscillations continue indefinitely, and the amplitude of the oscillations remains constant.

Hint

(b) The power factor of choke coil is given by
\(
\cos \phi=\frac{R}{\sqrt{R^2+\omega^2 L^2}} \approx \frac{R}{\omega L} \quad(\text { as } R \ll \omega L)
\)
As \(R \ll \omega L, \cos \phi\) is very small. Thus, the power absorbed by the coil is very small. The only loss of energy is due to hysteresis in the iron core, which is much less than the loss of energy in the resistance.

Hint

(c) \(X_C=\frac{1}{\omega C}=\frac{1}{2 \pi f C}\)
\(
X_C \propto \frac{1}{f}
\)
Therefore, \(X_C-f\) curve represents a hyperbola.

Hint

(b) By introducing the slab, \(C\) will increase. Therefore, \(X_C\) will decrease and new current will be more than \(I\).

Hint

\(
\begin{aligned}
&\text { (b) Applied voltage, } V=\sqrt{V_R^2+V_L^2}\\
&\therefore \quad V=\sqrt{(200)^2+(150)^2}=250 \mathrm{~V}
\end{aligned}
\)

Hint

(b)

\(
\begin{aligned}
I & =\frac{V}{Z}=\frac{V}{\sqrt{R^2+(2 \pi f L)^2}} \\
& =\frac{120}{\sqrt{(10)^2+(2 \pi \times 60 \times 2)^2}}=0.16 \mathrm{~A}
\end{aligned}
\)

Hint

(b) \(I_{avg}=5+0=5 \mathrm{~A}\)

Hint

(b)

\(
\begin{aligned}
E & =10 \cos (2 \pi f t)=10 \cos \left[2 \pi \times 50 \times \frac{1}{600}\right] \\
& =10 \cos \frac{\pi}{6}=5 \sqrt{3} \mathrm{~V}
\end{aligned}
\)

Hint

(c) Phase difference between current and voltage is \(\pi / 2\), with voltage leading, so \(X_L>X_C\) and \(R=0\).

Hint

(c) Resonant frequency, \(f=\frac{1}{2 \pi \sqrt{L C}}\)
It does not depend on resistance.

Hint

(a) Resonant frequency, \(f=\frac{1}{2 \pi \sqrt{L C}}\)
\(
\begin{aligned}
& f \propto \frac{1}{\sqrt{C}} \text { or } \frac{f_1}{f_2}=\sqrt{\frac{C_2}{C_1}} \\
& \frac{f}{f_2}=\sqrt{\frac{4 C}{C}} \Rightarrow f_2=\frac{f}{2}
\end{aligned}
\)

Hint

(c) Current in inductance \(I_L\) lags behind the voltage by \(\pi / 2\).
Current in resistance \(I_R\) is in phase with voltage, while current in capacitance \(I_C\) leads the voltage by a phase of \(\pi / 2\).

Hint

\(
\begin{aligned}
&\text { (b) Power dissipated in circuit, }\\
&P=\frac{V_{\mathrm{rms}}^2}{R}=\frac{(30)^2}{10}=90 \mathrm{~W}
\end{aligned}
\)

Hint

(a) In the circuit, voltage \(V=V_0 \sin \omega t\) and current
\(
I=T_0 \sin \left(\omega t-\frac{\pi}{2}\right)
\)
Phase different between voltage and current at any time would be \(\phi=\frac{\pi}{2}\)
So power consumed,
\(
\begin{aligned}
& P=\frac{I_0 V_0}{2} \cos \phi \\
& =\frac{I_0 V_0}{2} \cos \left(\frac{\pi}{2}\right) \\
& =P=0
\end{aligned}
\)

Hint

(c) \(\frac{H_{\mathrm{DC}}}{H_{\mathrm{AC}}}=\left(\frac{i_{\mathrm{DC}}}{i_{\mathrm{rms}}}\right)^2=\left(\frac{2}{\sqrt{2}}\right)^2=2: 1\)
\(\left(\because H \propto i^2\right)\)

Hint

(c)

\(
\begin{aligned}
I_0 & =28 \mathrm{~A} \\
I_{\mathrm{rms}} & =\frac{I_0}{\sqrt{2}} \approx 20 \mathrm{~A}
\end{aligned}
\)

Hint

(b)

\(
\begin{aligned}
P & =\frac{1}{2} V_0 I_0 \cos \phi \\
P & =P_{\text {peak }} \cos \phi \\
\frac{1}{2}\left(P_{\text {peak }}\right) & =P_{\text {peak }} \cos \phi \\
\cos \phi & =\frac{1}{2} \\
\phi & =\frac{\pi}{3}
\end{aligned}
\)

Hint

(a)

\(
\begin{aligned}
P & =V_{\mathrm{rms}} I_{\mathrm{rms}} \cos \phi \\
& =V_{\mathrm{rms}}\left(\frac{V_{\mathrm{rms}}}{Z}\right)\left(\frac{R}{Z}\right) \\
& =\frac{(110)^2(11)}{(22)^2}=275 \mathrm{~W}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (a) Here, } \quad V^2=V_R^2+V_L^2\\
&\therefore \quad V_L=\sqrt{V^2-V_R^2}=\sqrt{\left(20^2-(12)^2\right.}=16 \mathrm{~V}
\end{aligned}
\)

Hint

(d) The given voltage \(e=e_1 \sin \omega t+e_2 \cos \omega t\) can be written as \(e=E_0 \sin (\omega t+\phi)\).
The peak value \(E_0\) is given by \(E_0=\sqrt{e_1^2+e_2^2}\).
The RMS value of a sinusoidal voltage is its peak value divided by \(\sqrt{2}\).
\(
e_{r m s}=\frac{E_0}{\sqrt{2}}
\)
Substitute the expression for \(E_0: e_{r m s}=\frac{\sqrt{e_1^2+e_2^2}}{\sqrt{2}}\).

Hint

\(
\begin{aligned}
&\text { (c) Average value of output current is given by }\\
&\begin{aligned}
I_{\mathrm{av}} & =\frac{\int_0^{T / 2} d t}{\int_0^{T / 2} d t}=\frac{\int_0^{T / 2} I_0 \sin \omega t d t}{T / 2} \\
& =\frac{2 I_0}{T}\left[-\frac{\cos \omega t}{\omega}\right]_0^{T / 2} \\
& =\frac{2 I_0}{T}\left[-\frac{\cos (\omega T / 2)}{\omega}+\frac{\cos 0^{\circ}}{\omega}\right] \\
& =\frac{2 I_0}{\omega T}\left[-\cos \pi+\cos 0^{\circ}\right]=\frac{2 I_0}{2 \pi}[1+1]=\frac{2 I_0}{\pi}
\end{aligned}
\end{aligned}
\)

Hint

(c) Phase different between current and voltage,
\(
\Delta \phi=\phi_2-\phi_1=\frac{\pi}{6}-\left(-\frac{\pi}{6}\right)=\frac{\pi}{3} \text { or } 60^{\circ} .
\)
Hence, current leads voltage by \(60^{\circ}\).

Hint

\(
\begin{aligned}
& \text { (b) Reading of ammeter }=i_{\text {rms }} \\
& \qquad \begin{aligned}
i_{\text {rms }} & =\frac{V_{\text {rms }}}{X_C}=\frac{V_0}{\sqrt{2}} \times \omega C \\
& =\frac{200 \sqrt{2} \times 100 \times\left(1 \times 10^{-6}\right)}{\sqrt{2}} \\
& =2 \times 10^{-2} \mathrm{~A}=20 \mathrm{~mA}
\end{aligned}
\end{aligned}
\)

Hint

(a) In a series RLC circuit, current is maximum at resonance.
Resonance occurs when inductive reactance \(X_L\) equals capacitive reactance \(X_C\).
\(
\begin{aligned}
& X_L=\omega L=2 \pi n L \\
& X_C=\frac{1}{\omega C}=\frac{1}{2 \pi n C}
\end{aligned}
\)
At series resonance, \(2 \pi n L=\frac{1}{2 \pi n C}\).
The series resonance frequency is \(n=\frac{1}{2 \pi \sqrt{L C}}\).
In a parallel LC circuit, current is minimum at resonance (anti-resonance).
This also occurs when \(X_L=X_C\).
Let the parallel resonance frequency be \(n^{\prime}\).
\(
2 \pi n^{\prime} L=\frac{1}{2 \pi n^{\prime} C}
\)
The parallel resonance frequency is \(n^{\prime}=\frac{1}{2 \pi \sqrt{L C}}\).
The series resonance frequency \(n\) is \(\frac{1}{2 \pi \sqrt{L C}}\). The parallel resonance frequency \(n^{\prime}\) is \(\frac{1}{2 \pi \sqrt{L C}}\). Therefore, \(n^{\prime}=n\).

Hint

(d) \(\tan \phi=\frac{X_L}{R}\) (In case of \(L-R\) circuit)
\(
=\frac{L \omega}{R}=\frac{L 2 \pi f}{R}=\frac{2 \times 200}{300}=\frac{4}{3}
\)

Hint

\(
\text { (a) } \omega=\frac{1}{\sqrt{L C}}
\)
\(
\begin{aligned}
f & =\frac{1}{2 \pi \sqrt{L C}} \\
& =\frac{1}{2 \pi \sqrt{20 \times 10^{-6} \times 10 \times 10^{-3}}}=356 \mathrm{~Hz} \text { or cycles } / \mathrm{s}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { (b) } I_{\mathrm{rms}}=\sqrt{\left(I^2\right)}=\sqrt{(8+6 \sin \omega t)^2} \\
& \Rightarrow I_{\mathrm{rms}}=\sqrt{\left(\left(64+96 \sin \omega t+36 \sin ^2 \omega t\right)\right)} \\
& \Rightarrow I_{\mathrm{rms}}=\sqrt{(64)+96<\sin \omega t>+36<\sin ^2 \omega t>} \\
& \Rightarrow I_{\mathrm{rms}}=\sqrt{64+0+36 \times 0.5}=9.05 \mathrm{~A} \text { Since, }\left\langle\sin ^2 \omega t\right\rangle=0.5
\end{aligned}
\)

Hint

(d) Reactance of series L-C circuit \(=X_L+X_C\)
\(
=j\left(\omega L-\frac{1}{\omega C}\right)
\)
So, its value is infinite at both zero and infinite frequency. It is shown in graph (d).

Hint

\(
\begin{aligned}
&\text { (b) Value of induced emf in coil, }\\
&\begin{aligned}
e & =M \frac{d I}{d t} \\
& =0.005 \times \frac{d}{d t}\left(I_0 \sin \omega t\right) \\
& =0.005 \times I_0 \omega \cos \omega t \\
\therefore \quad e_{\max } & =0.005 \times 10 \times 100 \pi \\
& =5 \pi \mathrm{~V}
\end{aligned}
\end{aligned}
\)

Hint

(c) For only a resistance, it hardly matters, whether the source is DC or AC . In series,
\(
P=\frac{P_1 P_2}{P_1+P_2}=\frac{1000 \times 1000}{2000}=500 \mathrm{~W}
\)

Hint

\(
\begin{aligned}
& \text { (b) } X_L=R \\
& \therefore \quad Z=\sqrt{2} R
\end{aligned}
\)
\(
P=\left(\frac{V_{\mathrm{rms}}}{Z}\right)^2 \cdot R=\left(\frac{E_0 / \sqrt{2}}{\sqrt{2} R}\right)^2 \cdot R=\frac{E_0^2}{4 R}
\)

Hint

\(
\text { (b) Induced emf in the circuit, } e=M \frac{d i}{d t}=(0.2)(5)=1 \mathrm{~V}
\)

Hint

\(
\begin{aligned}
& \text { (b) } \frac{N_P}{N_S}=\frac{V_P}{V_S}=\frac{i_S}{i_P} \\
& \therefore \quad i_S=\frac{N_P i_P}{N_S}=\frac{4 \times 140}{280}=2 \mathrm{~A}
\end{aligned}
\)

Hint

(d) Here, \(I_R=I_L=I\)
and they have a phase difference \(0<\phi<90^{\circ}\), with applied voltage.

Hint

\(
\begin{aligned}
& \text { (d) } X_L=2 \pi f L=2 \pi\left(\frac{50}{\pi}\right) \times 1=100 \Omega \\
& \quad X_C=\frac{1}{2 \pi f C}=\frac{1}{2 \pi\left(\frac{50}{\pi}\right) 20 \times 10^{-6}}=500 \Omega \\
& \text { Impedance, } \quad Z=\sqrt{R^2+\left(X_C-X_L\right)^2} \\
& \Rightarrow \quad Z=\sqrt{(300)^2+(400)^2}=500 \Omega
\end{aligned}
\)

Hint

(c) As, emf induced in coil is \(e=-\frac{L d I}{d t}\)
\(\therefore\) Voltage versus time curve is a straight line with negative slope.

Hint

(a) Let the applied voltage be \(V\) volt

\(
\begin{aligned}
&\text { Here, }\\
&\begin{aligned}
V_R & =12 \mathrm{~V}, V_C=5 \mathrm{~V} \\
V & =\sqrt{V_R^2+V_C^2}=\sqrt{(12)^2+(5)^2} \\
& =\sqrt{144+25} \\
& =\sqrt{169}=13 \mathrm{~V}
\end{aligned}
\end{aligned}
\)

Hint

\(
\text { (a) } V_{\mathrm{rms}}=\sqrt{V_{\mathrm{av}}}=\sqrt{\frac{1}{T} \int_0^T V^2 d t}=\sqrt{\frac{1}{T} \int_0^T 10^2 d t}=10 \mathrm{~V}
\)

Hint

\(
\begin{aligned}
&\text { (c) Peak voltage, } V_0=\sqrt{2} V_{\mathrm{rms}}=331 \mathrm{~V}\\
&\begin{array}{ll}
\text { and } & \omega=2 \pi f=100 \pi \\
\Rightarrow & V_L=331 \sin (100 \pi t)
\end{array}\\
&\text { where, } V_L \text { is line voltage. }
\end{aligned}
\)

Hint

(d) In pure inductor, voltage leads the current. So, correct graph is (d).

Hint

\(
\begin{aligned}
&\text { (b) }\\
&\begin{aligned}
& \text { For capacitive circuits, } X_C=\frac{1}{\omega C} \\
& \therefore \quad I=\frac{V}{X_C}=V \omega C \Rightarrow I \propto \omega
\end{aligned}
\end{aligned}
\)

Hint

(d) Time taken by the current to reach the maximum value.
\(
\begin{aligned}
t & =\frac{T}{4}=\frac{1}{4 f}=\frac{1}{4 \times 50}=5 \times 10^{-3} \mathrm{~s} \\
I_0 & =\sqrt{2} I_{\mathrm{rms}}=10 \sqrt{2}=14.14 \mathrm{~A}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (d) According to problem, } X_L=X_C\\
&\begin{array}{ll}
& \omega L=\frac{1}{\omega C} \\
\therefore & \omega^2=\frac{1}{L C}=\frac{1}{\sqrt{10^{-3} \times 10 \times 10^{-6}}} \\
& \omega=\frac{1}{\sqrt{10^{-8}}}=10^4 \\
\therefore & X_L=\omega L=10^4 \times 10^{-3}=10 \Omega
\end{array}
\end{aligned}
\)

Hint

(c) At resonance frequency, current is maximum.

Hint

\(
\begin{aligned}
& \text { (b) } P=V_{\mathrm{rms}} I_{\mathrm{rms}}=\frac{V_0}{\sqrt{2}} I_{\mathrm{rms}} \\
& \therefore \quad I_{\mathrm{rms}}=\frac{1000 \sqrt{2}}{200}=5 \sqrt{2} \mathrm{~A}
\end{aligned}
\)

Hint

\(
\text { (a) } V=30 \sin 50 t+40 \cos 50 t=50 \sin \left(50 t+53^{\circ}\right)
\)

\(
\therefore \quad I_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{Z}=\frac{50 / \sqrt{2}}{10}=\frac{5}{\sqrt{2}} \mathrm{~A}
\)

Hint

\(
\begin{aligned}
&\begin{aligned}
& \text { (c) } I_0=\frac{140}{10}=14 \mathrm{~A} \\
& \therefore \quad I_{\mathrm{rms}}=\frac{I_0}{\sqrt{2}} \approx 10 \mathrm{~A}
\end{aligned}\\
&\therefore I_{\mathrm{DC}} \text { required will be approximately } 10 \mathrm{~A} \text {. }
\end{aligned}
\)

Hint

(c)

\(
\begin{aligned}
\begin{aligned}
i_{\mathrm{rms}}^2 & =\frac{\int i^2 d t}{\int d t} \\
& =\frac{\int_2^4(4 t) d t}{\int_2^4 d t}=\frac{4 \int_2^4 t d t}{2}=2\left[\frac{t^2}{2}\right]_2^4=12 \mathrm{~A}^2 \\
\Rightarrow i_{\mathrm{rms}} & =2 \sqrt{3} \mathrm{~A}
\end{aligned}
\end{aligned}
\)

Hint

(b) Power factor, \(\cos \phi=\frac{1}{\sqrt{2}}=\frac{R}{\sqrt{R^2+\omega^2 L^2}}\)
This gives, \(\omega L=R\)
When \(\omega\) is doubled, we get
\(
\cos \phi^{\prime}=\frac{R}{\sqrt{R^2+4 \omega^2 L^2}}=\frac{R}{\sqrt{R^2+4 R^2}}=\frac{1}{\sqrt{5}}
\)

Hint

(d) Resistance of coil \((R)=\frac{200}{1}=200 \Omega\)
With AC source, \(I=\frac{200}{\sqrt{R^2+X_L^2}} \quad\) or \(\quad 0.5=\frac{200}{\sqrt{R^2+X_L^2}}\)
\(
\begin{aligned}
R^2+(2 \pi f L)^2 & =(400)^2 \\
\left(2 \pi f \times \frac{2 \sqrt{3}}{\pi}\right)^2 & =(400)^2-(200)^2=200 \times 600 \\
4 \sqrt{3} f & =\sqrt{12} \times 100 \Rightarrow f=50 \mathrm{~Hz}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\begin{array}{ll}
\text { (a) From, } & P=V I=\frac{V^2}{R} \\
\Rightarrow & I=\frac{P}{V}=6 \mathrm{~A} \Rightarrow R=\frac{V^2}{P}=\frac{100}{60}=\frac{5}{3} \Omega \\
\text { Now, } & I=\frac{V}{Z} \\
\Rightarrow & 6=\frac{100}{\sqrt{\left(\frac{5}{3}\right)^2+(2 \pi f L)^2}}
\end{array}\\
&\text { Solving this equation, we get }\\
&L=0.052 \mathrm{H}
\end{aligned}
\)

Hint

(a) Given, \(X_L=X_C=5 \Omega\), this is the condition of resonance. So \(V_L=V_C\), or net voltage across \(L\) and \(C\) combination will be zero.
\(
\begin{aligned}
\therefore \quad Z & =R=55 \Omega \\
I & =\frac{V}{Z}=\frac{110}{55}=2 \mathrm{~A}
\end{aligned}
\)

Hint

(d) Since, \(V_C>V_L\)
\(
\therefore \quad X_C>X_L
\)
Hence, current will lead the voltage.
\(
\begin{aligned}
V & =\sqrt{V_R^2+\left(V_C-V_L\right)^2}=10 \mathrm{~V}<V_C \\
\cos \phi & =\frac{R}{Z}=\frac{V_R}{V}=\frac{8}{10}=\frac{4}{5}
\end{aligned}
\)

Hint

\(
\begin{array}{ll}
\text { (a) Impedance, } Z=\sqrt{R^2+\left(X_L-X_C\right)^2} \\
\therefore 10=\sqrt{(10)^2+\left(X_L-X_C\right)^2} \\
\Rightarrow 100=100+\left(X_L-X_C\right)^2 \\
\Rightarrow X_L-X_C=0 \dots(i)
\end{array}
\)
Let \(\phi\) is the phase difference between current and voltage,
\(
\therefore \quad \tan \phi=\frac{X_L-X_C}{R}
\)
\(
\tan \phi=\frac{0}{R} \quad \text { [from Eq. (i)] }
\)
\(
\phi=0
\)

Hint

(b) Resonance frequency,
\(
\begin{aligned}
& y, \omega=\frac{1}{\sqrt{L C}} \\
& =\frac{1}{\sqrt{8 \times 10^{-3} \times 20 \times 10^{-6}}} \\
& =2500 \mathrm{rads}^{-1}
\end{aligned}
\)
Resonance current, \(I=\frac{V}{R}=\frac{220}{44}=5 \mathrm{~A}\)

Hint

(c) At point \(A, \quad X_C>X_L\)
at point \(B, \quad X_C=X_L\)
at point \(C, \quad X_C<X_L\)
So at \(C\), circuit is inductive.

Hint

\(
\begin{aligned}
& \text { (d) At resonance, } X_L=X_C \\
& \therefore \quad Z=R=40 \Omega \\
& \text { and } \quad I_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{Z}=\frac{230}{40}=5.75 \mathrm{~A}
\end{aligned}
\)

Hint

(b) At resonance,
\(
\begin{aligned}
& & X_L & =X_C \\
\therefore & & V_1 & =V_3
\end{aligned}
\)
Hence, \(V_2\) will be zero.

Hint

(a)

\(
\begin{aligned}
Z & =\sqrt{(52)^2+\left[230 \times 10^{-3} \times 2 \pi \times 80-\frac{1}{2 \pi \times 80 \times 8.8 \times 10^{-6}}\right]^2} \\
& =122 \Omega \\
P & =\left(\frac{V_{\mathrm{rms}}}{Z}\right)^2 \cdot R=\left(\frac{150}{122}\right)^2(52)=78.6 \mathrm{~W}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (a) } C_{\text {net }}=C / 2 \text { (in series) and } L_{\text {net }}=2 L \text { (in series) }\\
&\begin{aligned}
& \therefore \quad \omega=\frac{1}{\sqrt{L_{\text {net }} C_{\text {net }}}} \\
& \text { or } \quad f=\frac{1}{2 \pi \sqrt{(2 L) \cdot\left(\frac{C}{2}\right)}} \text { or } f=\frac{1}{2 \pi \sqrt{L C}}
\end{aligned}
\end{aligned}
\)

Hint

(c) Four resistances form a balanced Wheatstone bridge. Current flowing in the coil will be zero and hence energy stored in the coil is zero.

Hint

\(
\begin{aligned}
&\begin{aligned}
& \text { (b) } N_P=100, P_i=10 \mathrm{~kW}, V_i=200 \mathrm{~V} \\
& \therefore I_P=\frac{P_i}{V_i}=\frac{10000}{200}=50 \mathrm{~A}
\end{aligned}\\
&\text { Input power }=\text { Output power }(\text { Neglecting all losses })\\
&\therefore \quad I_S=\frac{P_i}{V_o}=\frac{10000}{5000}=2 \mathrm{~A}
\end{aligned}
\)

Hint

\(
\begin{array}{ll}
\text { (c) } R=X_L=\frac{V_o}{I_o}=\frac{200}{5}=40 \Omega \\
\therefore Z=\sqrt{R_f^2+X_L^2}=40 \sqrt{2} \Omega \\
\Rightarrow I_{\text {rms }}=\frac{V_{\text {rms }}}{Z}=\frac{200 / \sqrt{2}}{40 \sqrt{2}}=\frac{5}{2} \mathrm{~A}
\end{array}
\)

Hint

(c) Let \(R\) be the resistance of coil and capacitor. Then,
\(
R=\frac{24}{6}=4 \Omega
\)
In the second case,
\(
I_{\max }=\frac{12}{4+4}=\frac{12}{8}=1.5 \mathrm{~A}
\)

Hint

\(
\begin{aligned}
&\text { (d) } X_C>X_L \text {. Hence, current will lead the voltage. }\\
&Z=\sqrt{R^2+\left(X_C-X_L\right)^2}=10 \sqrt{2} \Omega
\end{aligned}
\)
\(
\begin{aligned}
& I_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{Z}=\frac{400 / \sqrt{2}}{10 \sqrt{2}}=20 \mathrm{~A} \\
& \cos \phi=\frac{R}{Z}=\frac{1}{\sqrt{2}} \\
& V_R=I_{\mathrm{rms}} R=(20)(10)=200 \mathrm{~V}
\end{aligned}
\)

Hint

(d) Taking, \(V^2=V_R^2+\left(V_L-V_C\right)^2\)
According to the figure, we conclude that
\(
V_L=V_C
\)
\(
\begin{aligned}
& \therefore \quad V_R=V=220 \mathrm{~V} \\
& \text { Reading of voltmeter }=220 \mathrm{~V} \\
& \text { Reading of ammeter, } I=\frac{220}{100}=2.2 \mathrm{~A}
\end{aligned}
\)

Hint

(d) The voltages \(V_L\) and \(V_C\) are equal and opposite, so, voltmeter reading will be zero.
Also, \(R=30 \Omega, X_L=X_C=25 \Omega\)
So,
\(
\begin{aligned}
I & =\frac{V}{\sqrt{R^2+\left(X_L-X_C\right)^2}} \\
& =\frac{V}{R}=\frac{240}{30}=8 \mathrm{~A}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (b) Resonance frequency, }\\
&\omega_0=\frac{1}{\sqrt{L C}}=\frac{1}{\sqrt{100 \times 10^{-6} \times 1 \times 10^{-6}}}=10^5 \mathrm{rads}^{-1}
\end{aligned}
\)
\(
\begin{aligned}
& \text { Now, given } \omega=70 \text { kilo-rads }^{-1}=70000 \text { rads }^{-1} \\
& \Rightarrow \quad \omega_0>\omega
\end{aligned}
\)
\(
\begin{aligned}
&\begin{array}{ll}
\text { As } & X_L \propto \omega \text { and } X_C \propto \frac{1}{\omega} \\
\therefore & X_L>X_C
\end{array}\\
&\text { The circuit will be series } R \text { – } L \text { circuit. }
\end{aligned}
\)

Hint

\(
\begin{aligned}
\text { (b) } I_L= & I_{O L} \sin \left(\omega t-\frac{\pi}{2}\right)=-1.2 \cos \omega t \\
& I_C=I_{O C} \sin \left(\omega t+\frac{\pi}{2}\right)=1.0 \cos \omega t \\
\therefore & I=I_L+I_C=-0.2 \cos \omega t \Rightarrow|I|=0.2 \mathrm{~A}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (b) Voltage drop across } R \text { is maximum at resonance, i.e. at }\\
&\begin{aligned}
& \omega=\frac{1}{\sqrt{L C}} \text { or } f=\frac{\omega}{2 \pi}=\frac{1}{2 \pi \sqrt{L C}} \\
& =\frac{1}{2 \pi \sqrt{\left(\frac{1}{\pi}\right)\left(\frac{1}{\pi} \times 10^{-6}\right)}}=500 \mathrm{~Hz}
\end{aligned}
\end{aligned}
\)

Hint

(d) Given, \(I_1=I_2\)
\(
\therefore \quad \omega_1 L-\frac{1}{\omega_1 C}=\frac{1}{\omega_2 C}-\omega_2 L
\)
Solving this, we get
\(
\frac{1}{\sqrt{L C}}=\text { resonance frequency }=\sqrt{\omega_1 \omega_2}=\sqrt{9 \times 4}=6 \mathrm{kHz}
\)
At resonance frequency, current will be maximum.

Hint

(a) The given AC circuit is the combination of two pure parallel circuits with the applied voltage. In this, \(I_2\) is in phase with \(V\) and \(I_1\) leads \(V\) by \(90^{\circ}\).

Hint

(d) At resonance, Voltage across \(L=\) Voltage across \(C\)
\(\therefore \quad\) Reading in \(V_2=\) Reading in \(V_3\).

Hint

(b) \(I_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{\sqrt{R^2+\frac{1}{\omega^2 C^2}}}\)
When \(\omega\) increases, \(I_{\text {rms }}\) increases, so the bulb glows brighter.

Hint

(a) Total resistance of the circuit, \(R=6+4=10 \Omega\) Capacitive reactance, \(X_C=\frac{1}{\omega C}=\frac{1}{2000 \times 50 \times 10^{-6}}=10 \Omega\)
Inductive reactance,
\(
\begin{array}{ll}
& X_L=\omega L=2000 \times 5 \times 10^{-3}=10 \Omega \\
\therefore & Z=\sqrt{R^2+\left(X_L-X_C\right)^2}=10 \Omega
\end{array}
\)
Amplitude of current, \(I_0=\frac{V_0}{Z}=\frac{20}{10}=2 \mathrm{~A}\)

Hint

\(
\begin{aligned}
&\text { (a) }\\
&\begin{aligned}
X_C & =X_L=R \\
V & =\sqrt{V_R^2+\left(V_L-V_C\right)^2}
\end{aligned}
\end{aligned}
\)
\(
\begin{aligned}
&\begin{array}{ll}
\text { but } & V_L=V_C \\
\therefore & V=V_R=10 \mathrm{~V}
\end{array}\\
&\text { When capacitor is short circuited, } V_C=0
\end{aligned}
\)
\(
\text { and } \quad V_R=V_L \quad\left(\text { as } R=X_L\right)
\)
\(
\begin{aligned}
V & =\sqrt{V_R^2+V_L^2} \quad \text { or } 10=\sqrt{2} V_L \\
V_L & =\frac{10}{\sqrt{2}} \mathrm{~V}
\end{aligned}
\)

Hint

(d) \(\quad V_C=200 \mathrm{~V} \text { [given] }\)
\(
\begin{aligned}
V_R & =(11)(20)=220 \mathrm{~V} \\
V & =220 \mathrm{~V} \text { [applied] }
\end{aligned}
\)
\(
\begin{aligned}
V & =\sqrt{V_R^2+\left(V_L-V_C\right)^2} \\
V_L & =V_C=200 \mathrm{~V}
\end{aligned}
\)

Hint

(c) Given, \(X_L=1 \Omega, R=2 \Omega\)
\(
V_{\mathrm{rms}}=6 \mathrm{~V}, P_{\mathrm{av}}=?
\)
Average power dissipated in the circuit,
\(
\begin{aligned}
P_{\mathrm{av}} & =V_{\mathrm{rms}} I_{\mathrm{rms}} \cos \phi \dots(i) \\
I_{\mathrm{rms}} & =\frac{I_0}{\sqrt{2}}=\frac{V_{\mathrm{rms}}}{Z} \\
Z & =\sqrt{R^2+X_L^2}=\sqrt{4+1}=\sqrt{5} \Omega \\
I_{\mathrm{rms}} & =\frac{6}{\sqrt{5}} \mathrm{~A} \\
\cos \phi & =\frac{R}{Z}=\frac{2}{\sqrt{5}} \\
P_{\mathrm{av}} & =6 \times \frac{6}{\sqrt{5}} \times \frac{2}{\sqrt{5}} \text { [from Eq. (i)] }\\
& =\frac{72}{\sqrt{5} \sqrt{5}}=\frac{72}{5}=14.4 \mathrm{~W}
\end{aligned}
\)

Hint

(b) \(X_L=(2 \pi f L)=2 \pi \times 50 \times 0.7=220 \Omega\)
\(
\begin{aligned}
\therefore \quad Z & =\sqrt{R^2+X_L^2}=220 \sqrt{2} \Omega \\
I_{\mathrm{rms}} & =\frac{V_{\mathrm{rms}}}{Z}=\frac{220}{220 \sqrt{2}}=\frac{1}{\sqrt{2}} \mathrm{~A}
\end{aligned}
\)
\(\therefore\) Power, \(\quad P=I_{\mathrm{rms}}^2 \cdot R=\frac{1}{2} \times 220=110 \mathrm{~W}\)

Hint

\(
\text { (d) } V_L=V_C \Rightarrow X_L=X_C \text { or circuit shown is in resonance. }
\)
\(
\begin{aligned}
V_{\text {applied }} & =V_R=100 \mathrm{~V} \\
I & =\frac{V_R}{R}=2 \mathrm{~A}
\end{aligned}
\)

Hint

(d) \(V_S^2=(3)^2+(8-4)^2 \Rightarrow V_S=5 \mathrm{~V}\)
Now, \(\quad Z=\frac{V_S}{I}=\frac{5}{1}=5 \Omega\)
Also, \(\quad V_R=I R\) or \(R=\frac{3}{1}=3 \Omega\)
So, \(\quad \mathrm{PF}=\frac{R}{Z}=\frac{3}{5}=0.6 ;\) Also \(V_L>V_C\)
As, \(I\) lags \(V\), so PF is lagging in nature.

Hint

(d) Phase angle, \(\tan \phi=\frac{X_L}{R}=\frac{X_C}{R}\)
\(
\begin{array}{ll}
\Rightarrow & \tan 60^{\circ}=\frac{X_L}{R}=\frac{X_C}{R} \Rightarrow X_L=X_C=\sqrt{3} R \\
\text { i.e., } & Z=\sqrt{R^2+(\sqrt{3} R-\sqrt{3} R)^2} \Rightarrow Z=R
\end{array}
\)
So, average power, \(P=\frac{V^2}{R}=\frac{200 \times 200}{100}=400 \mathrm{~W}\)

Hint

(b) Power consumed in the coil,
\(
\begin{aligned}
& P=I_{\mathrm{rms}}^2 R \\
& R=\frac{P}{I_{\mathrm{rms}}^2}=\frac{240}{16}=15 \Omega \\
& \text { Impedance, } \quad Z=\frac{V}{I}=\frac{100}{4}=25 \Omega \\
& \text { Now, } \\
& X_L=\sqrt{Z^2-R^2} \\
& =\sqrt{(25)^2-(15)^2}=20 \Omega \\
& \therefore \quad 2 \pi f L=20 \\
& L=\frac{20}{2 \pi \times 50}=\frac{1}{5 \pi} \mathrm{H}
\end{aligned}
\)

Hint

(b) In an L-C-R circuit, the current and the voltage are in phase ( \(\phi=0\) ), when
\(
\begin{aligned}
& \quad \tan \phi=\frac{\omega L-\frac{1}{\omega C}}{R}=0 \text { or } \omega L=\frac{1}{\omega C} \\
& \text { or } \quad L=\frac{1}{\omega^2 C} \\
& \text { Here, } \quad \omega=2 \pi f=2 \times 3.14 \times 50 \mathrm{~s}^{-1}=314 \mathrm{~s}^{-1} \\
& \text { and } \quad C=20 \mu \mathrm{~F}=20 \times 10^{-6} \mathrm{~F} \\
& \therefore \quad L=\frac{1}{(314)^2 \times\left(20 \times 10^{-6}\right)}=0.51 \mathrm{H}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (a) Angular velocity, }\\
&\begin{aligned}
\omega_0 & =2 \pi v=2 \pi \times 100 \\
\omega_0 & =2 \times 3 \times 100 \\
& =600 \mathrm{rads}^{-1}
\end{aligned}\\
&[\because \pi=3]
\end{aligned}
\)
Further, \(\omega_0=\frac{1}{\sqrt{L C}} \dots(i)\)
Also, \(X_C=\frac{1}{C \omega_0}=60 \Omega\)
\(
\begin{aligned}
& C=\frac{1}{\omega_0 \times 60}=\frac{1}{600 \times 60} \\
& C=\frac{1}{36 \times 10^3} \mathrm{~F}
\end{aligned}
\)
Putting values of \(C\) in Eq. (i), we get
\(
\begin{aligned}
600 & =\frac{1}{\sqrt{L\left(\frac{1}{36 \times 10^3}\right)}} \\
36 \times 10^4 & =\frac{36 \times 10^3}{L} \\
L & =\frac{36 \times 10^3}{36 \times 10^4}=\frac{1}{10}=0.1 \mathrm{H}
\end{aligned}
\)

Hint

(a) Charge on the capacitor,
\(
q_0=C V=1 \times 10^{-6} \times 1=10^{-6} \mathrm{C}
\)
Here, \(q=q_0 \sin \omega t\)
or \(I_0=\omega q_0=\text { maximum current }\)
Now, \(\omega=\frac{1}{\sqrt{L C}}=\frac{1}{\sqrt{10^{-9}}}=\left(10^9\right)^{1 / 2}\)
\(
\therefore \quad I_0=\left(10^9\right)^{1 / 2} \times\left(1 \times 10^{-6}\right)=\sqrt{1000} \mathrm{~mA}
\)

Hint

\(
\begin{aligned}
&\text { (b) Circuit is in resonance, }\\
&\begin{array}{ll}
\therefore & V=V_R=I R \\
\text { or } & R=\frac{V}{I}=\frac{100}{0.2}=500 \Omega \\
& X_L=\frac{V_L}{I}=\frac{400}{0.2}=2000 \Omega=\omega L \\
\therefore & \omega=1000 \mathrm{rads}^{-1}
\end{array}
\end{aligned}
\)
\(
\begin{array}{ll}
\text { Further, } & X_L=X_C=2000 \Omega=\frac{1}{\omega C} \\
\therefore & C=\frac{1}{2000 \omega}=0.5 \mu \mathrm{~F}
\end{array}
\)

Hint

\(
\text { (a) Impedance }=\frac{V_{\mathrm{AC}}}{I_{\mathrm{AC}}}=\frac{100}{0.5}=200 \Omega=Z
\)
\(
\begin{aligned}
R & =\frac{V_{\mathrm{DC}}}{I_{\mathrm{DC}}}=\frac{100}{1}=100 \Omega \\
X_L & =\sqrt{Z^2-R^2}=100 \sqrt{3} \Omega=(2 \pi f L) \\
L & =\frac{100 \sqrt{3}}{2 \pi f}=\frac{100 \times 1.732}{2 \times 3.14 \times 50}=0.55 \mathrm{H}
\end{aligned}
\)

Hint

(c) \(X_L=\frac{100}{8} \Omega=12.5 \Omega\) with 50 Hz frequency and with 40 Hz frequency.
\(
X_L^{\prime}=12.5 \times \frac{40}{50}=10 \Omega \quad\left[\text { as } X_L \propto \omega\right]
\)
\(
\begin{aligned}
& R=\frac{100}{10}=10 \Omega \\
& Z=\sqrt{R^2+\left(X_L^{\prime}\right)^2}=10 \sqrt{2} \Omega \\
& I=\frac{100}{10 \sqrt{2}}=5 \sqrt{2} \mathrm{~A}
\end{aligned}
\)

Hint

\(
\text { (d) Let } \quad V=V_0 \sin \omega t \quad \text { [as } V=0 \text { at } t=0 \text { ] }
\)
\(
\text { Then } \quad V_R=V_0 \sin \omega t
\)
and \(\quad V_C=V_0 \sin (\omega t-\pi / 2)\)
\(V\) and \(V_R\) are in same phase. While \(V_C\) lags \(V\) (or \(V_R\) ) by \(90^{\circ}\).
Now \(V_R\) is in same phase with initial potential difference across the capacitor for the first time when,
\(
\begin{array}{ll}
& \omega t=-\frac{\pi}{2}+2 \pi=\frac{3 \pi}{2} \\
\therefore & t=\frac{3 \pi}{2 \omega}
\end{array}
\)

Hint

(b) At resonance, \(X_C=X_L\)
If \(\omega\) is reduced to half,
\(
X_C^{\prime}=2 X_C \quad\left[\text { as } X_C \propto \frac{1}{\omega}\right]
\)
and \(X_L^{\prime}=\frac{X_L}{2} \left[\text { as } X_L \propto \omega\right]\)
If \(\omega\) is made two times,
\(
X_C^{\prime}=\frac{X_C}{2}
\)
and \(X_L^{\prime}=2 X_L\)
i.e. Value of \(Z\) will remain unchanged and current should be 1 A.

Hint

(b) In the given graph current is leading the voltage by \(45^{\circ}\). Therefore, circuit is \(C-R\).
\(
\tan \phi=\frac{X_C}{R} \text { and } \phi=\frac{\pi}{4}
\)
\(
\begin{aligned}
&\begin{array}{ll}
\therefore & X_C=R \\
\Rightarrow & \omega C R=1 \\
\Rightarrow & C R=\frac{1}{\omega}=\frac{1}{100}
\end{array}\\
&\text { When } R=1 \mathrm{k} \Omega=10^3 \Omega \text {, then }\\
&C=\frac{1}{100 \times 10^3}=10^{-5} \mathrm{~F}=10 \mu \mathrm{~F}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (a) In position-1, }\\
&\tan \frac{\pi}{6}=\frac{\omega L}{R}=\frac{(1000)\left(\sqrt{3} \times 10^{-3}\right)}{R}
\end{aligned}
\)
\(
\begin{gathered}
\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{R} \\
R=3 \Omega
\end{gathered}
\)
\(
\begin{aligned}
&\text { In position-2, }\\
&\begin{aligned}
\tan \phi & =\frac{X_C}{R}=\frac{(1 / \omega C)}{R} \\
& =\frac{1 /\left(1000 \times \frac{1000}{3} \times 10^{-6}\right)}{3}=1 \\
\therefore \quad \phi & =\frac{\pi}{4} ; \text { so current leads source emf. }
\end{aligned}
\end{aligned}
\)

Hint

(b) For L-C, oscillator, \(T=2 \pi \sqrt{L C}\)
From one extreme to another; when electric field is maximum magnetic field is zero. And when electric field is zero, magnetic field is maximum.
At time \(t=\frac{T}{4}\), energy stored is completely magnetic.
\(
\begin{aligned}
&\therefore \quad \text { Time, } t=\frac{T}{4}=\frac{2 \pi \sqrt{L C}}{4}\\
&\begin{array}{ll}
\Rightarrow & t=\frac{\pi \sqrt{20 \times 10^{-3} \times 50 \times 10^{-6}}}{2}=\frac{\pi \sqrt{1 \times 10^{-6}}}{2} \\
\Rightarrow & t=1.57 \mathrm{~ms}
\end{array}
\end{aligned}
\)

Hint

(b) \(P\) is a capacitor, \(X_C=\frac{220}{0.25}=880 \Omega\)
\(Q\) is resistance, \(R=\frac{220}{0.25}=880 \Omega\)
When \(P\) and \(Q\) are in series.
\(
\begin{aligned}
& Z=\sqrt{R^2+X_C^2} \\
&=880 \sqrt{2} \Omega \\
& \phi=\tan ^{-1}\left(\frac{X_C}{R}\right)=\pi / 4 \quad \text { (current leading) } \\
& I=\frac{V}{Z}=\frac{1}{4 \sqrt{2}} \mathrm{~A}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { (c) } X_L=\omega L=(2000)\left(5 \times 10^{-3}\right)=10 \Omega \\
& \quad X_C=\frac{1}{\omega C}=\frac{1}{(2000)\left(50 \times 10^{-6}\right)}=10 \Omega
\end{aligned}
\)
Since, \(\quad X_C=X_L\)
\(
\therefore \quad I=\frac{V}{Z}=\frac{20 / \sqrt{2}}{10}=\sqrt{2} \mathrm{~A}
\)
or ammeter reading \(=\sqrt{2} \mathrm{~A}\) and voltmeter reading \(=0\)

Hint

(b) In L-C oscillation energy is transferred from \(C\) to \(L\) or \(L\) to \(C\). Maximum energy in \(L=\frac{1}{2} L I_{\max }^2\) and maximum energy in \(C=\frac{q_{\max }^2}{2 C}\).
Equal energy will be, when
\(
\begin{aligned}
\frac{1}{2} L I^2 & =\frac{1}{2}\left(\frac{1}{2} L I_{\max }^2\right) \\
I & =\frac{1}{\sqrt{2}} I_{\max } \\
I & =I_{\max } \sin \omega t=\frac{1}{\sqrt{2}} I_{\max }
\end{aligned}
\)
\(
\begin{array}{ll}
\Rightarrow & \quad \omega t=\frac{\pi}{4} \\
\text { or } & 2 \frac{\pi}{T} t=\frac{\pi}{4} \quad \text { or } t=\frac{T}{8} \\
\Rightarrow & \quad t=\frac{1}{8} 2 \pi \sqrt{L C}=\frac{\pi}{4} \sqrt{L C}
\end{array}
\)

Hint

(b) Resistance of circuit,
\(
R=\frac{V^2}{P}=\frac{110 \times 110}{330}=\frac{110}{3} \Omega
\)
Ist case Power factor, \(\cos \phi=0.6\)
Since, current lags the voltage, thus the circuit contains resistance and inductance.
\(
\begin{array}{lr}
\therefore & \cos \phi=\frac{R}{\sqrt{R^2+X_L^2}}=0.6 \\
\Rightarrow & R^2+X_L^2=\left(\frac{R}{0.6}\right)^2 \\
\Rightarrow & X_L^2=\frac{R^2}{(0.6)^2}-R^2 \\
\Rightarrow & X_L^2=\frac{R^2 \times 0.64}{0.36} \\
\therefore & X_L=\frac{0.8 R}{0.6}=\frac{4 R}{3} \dots(i)
\end{array}
\)
IInd case Now \(\cos \phi=1 \text { [given] }\)
Therefore, circuit is purely resistive, i.e. it contains only resistance. This is the condition of resonance in which
\(
\begin{array}{ll}
& X_L=X_C \\
\therefore & X_C=\frac{4 R}{3}=\frac{4}{3} \times \frac{110}{3}=\frac{440}{9} \Omega \quad \text { [from Eq. (i)] } \\
\Rightarrow & \frac{1}{2 \pi f C}=\frac{440}{9} \Omega \\
\therefore & C=\frac{9}{2 \times 3.14 \times 60 \times 440}=0.000054 \mathrm{~F}=54 \mu \mathrm{~F}
\end{array}
\)

Hint

\(
\text { (c) } R=\frac{80}{10}=8 \Omega
\)

Also,
\(
\begin{aligned}
V_L^2+80^2 & =220^2 \\
V_L^2 & =(220+80)(220-80) \\
& =300 \times 140
\end{aligned}
\)
\(
\begin{aligned}
V_L & =204.9 \mathrm{~V} \\
I_{\mathrm{rms}} X_L & =204.9 \\
\frac{220}{\sqrt{64+X_L^2}} X_L & =204.9 \\
X_L & =20 \Rightarrow L=\frac{20}{2 \pi f}=0.064 \mathrm{H}
\end{aligned}
\)

Hint

(\(
\begin{aligned}
&\text { (d) }\\
&\begin{aligned}
Z & =\sqrt{(R)^2+\left(X_L-X_C\right)^2} \\
R & =10 \Omega, X_L=\omega L=2000 \times 5 \times 10^{-3}=10 \Omega
\end{aligned}
\end{aligned}
\)
\(
\begin{aligned}
X_C & =\frac{1}{\omega C}=\frac{1}{2000 \times 50 \times 10^{-6}} \\
& =10 \Omega \Rightarrow Z=10 \Omega
\end{aligned}
\)
\(
\begin{aligned}
& \text { Maximum current, } i_0=\frac{V_0}{Z}=\frac{20}{10}=2 \mathrm{~A} \\
& \text { Hence, } i_{\text {rms }}=\frac{2}{\sqrt{2}}=1.4 \mathrm{~A} \\
& \text { and } V_{\text {rms }}=4 \times 1.4=5.6 \mathrm{~V}
\end{aligned}
\)

Hint

\(
\begin{array}{r}
\text { (b) } Z=\sqrt{R^2+\left(X_C-X_L\right)^2}=10 \Omega \\
\cos \phi=\frac{R}{Z}=\frac{4}{5}
\end{array}
\)
\(
\begin{aligned}
& \phi=37^{\circ} \\
& I_0=\frac{V_0}{Z}=\frac{10}{10}=1 \mathrm{~A} \\
& I=1 \cos \left(100 \pi t+37^{\circ}\right)
\end{aligned}
\)
Given, \(\quad V=10 \cos (100 \pi t)\)
The applied voltage is \(\frac{3}{5}\) th of the maximum applied voltage. Then, \(100 \pi t=\cos ^{-1}(3 / 5)=53^{\circ}\)
\(
\begin{aligned}
I & =1 \cos \left(53^{\circ}+37^{\circ}\right)=0 \\
V_R & =0 \\
V_{\mathrm{AB}} & =V_{\text {applied }}=\frac{3}{5} \times 10=6 \mathrm{~V}
\end{aligned}
\)

Hint

\(
\text { (a) At any time } t \text { apply KVL, } \frac{q}{C}-i R-L \frac{d i}{d t}=0
\)

\(
\begin{array}{lc}
\text { and } & i=-\frac{d q}{d t} \\
\Rightarrow & \frac{q}{C}+\frac{d q}{d t} R+\frac{L d^2 q}{d t^2}=0 \\
\Rightarrow & \frac{d^2 q}{d t^2}+\frac{R}{L} \frac{d q}{d t}+\frac{q}{L C}=0
\end{array}
\)
From damped harmonic oscillator, the amplitude is given by \(A=A_0 e^{-\frac{b t}{2 m}}\), for general equation of double differential
\(
\begin{array}{ll}
\text { equation } \frac{d^2 x}{d t^2}+\frac{b}{m} \frac{d x}{d t}+\frac{k}{m} x=0 \\
\Rightarrow & Q_{\max }=Q_0 e^{-\frac{R t}{2 L}} \\
\Rightarrow & Q_{\max }^2=Q_0^2 e^{-\frac{R t}{L}}
\end{array}
\)
Lesser the self-inductance, faster will be damping, hence correct graph is in option (a).

Hint

(b) Given,
\(
\begin{gathered}
L=25 \mathrm{mH}=25 \times 10^{-3} \mathrm{H} \\
C=10 \mu \mathrm{~F}=10^{-5} \mathrm{~F}
\end{gathered}
\)
If \(T\) be the time period of \(L-C\) oscillation, then
\(
\begin{aligned}
T & =2 \pi \sqrt{L C}=2 \pi \sqrt{25 \times 10^{-3} \times 10^{-5}} \\
& =\pi \times 10^{-3} \mathrm{~s}=\pi \mathrm{ms}
\end{aligned}
\)
Current in the circuit will be maximum, when
\(
t=\frac{T}{4}=\frac{\pi}{4} \mathrm{~ms}
\)

Hint

(a) Given, in \(L-C-R\) series circuit,
Source voltage (resultant voltage), \(V=120 \mathrm{~V}\)
Voltage across inductor, \(V_L=50 \mathrm{~V}\)
Voltage across resistor, \(V_R=40 \mathrm{~V}\)
In \(L-C-R\) series circuit,
\(
\begin{aligned}
V & =\sqrt{V_R^2+\left(V_L-V_C\right)^2} \\
120 & =\sqrt{40^2+\left(50-V_C\right)^2} \\
120^2 & =40^2+\left(50-V_C\right)^2 \\
\left(50-V_C\right)^2 & =12800 \\
50-V_C & =80 \sqrt{2} \\
V_C & =10(5-8 \sqrt{2})
\end{aligned}
\)

Hint

(b) At \(t \rightarrow \infty\), capacitor works as an open circuit as shown in the figure.
Therefore, current flow in the circuit,

\(
\begin{aligned}
I_1 & =\frac{9}{(12+15) \times 10^3}=\frac{1}{3} \times 10^{-3} \mathrm{~A} \\
\therefore \quad V_0 & =I_1 \times 15 \times 10^3 \\
& =\frac{1}{3} \times 10^{-3} \times 15 \times 10^3=5 \mathrm{~V}
\end{aligned}
\)
\(\therefore\) Charge on capacitor of \(5 \mu \mathrm{~F}\),
\(
\begin{aligned}
q_0 & =C V_0=5 \times 10^{-6} \times 5 \\
& =25 \times 10^{-6} \mathrm{C}=25 \mu \mathrm{C}
\end{aligned}
\)
When switch is open, then capacitor starts discharging across \((5+15=20 \mathrm{k} \Omega)\) and instantaneous charge on capacitor is given by
\(
\begin{aligned}
q & =q_0 e^{-t / R C} \\
& =25 e^{-\frac{1}{20 \times 10^3 \times 5 \times 10^{-6}}} [\because t=1 \mathrm{~s}]\\
& =25 e^{-10} \mu \mathrm{C}
\end{aligned}
\)

Hint

(a) Given, turn ratio of a transformer,
\(
\begin{aligned}
& & \frac{N_1}{N_2} & =\frac{50}{1} \\
\Rightarrow & & N_1 & =50 N_2
\end{aligned}
\)
Since, \(\frac{N_1}{N_2}=\frac{V_1}{V_2}\)
\(
50=\frac{120}{V_2} \quad\left[V_1=120 \mathrm{~V}(\text { given })\right]
\)
\(
\begin{aligned}
&\Rightarrow \quad V_2=\frac{12}{5} \mathrm{~V}\\
&\text { Output power at secondary coil, }\\
&\begin{aligned}
P_s & =\frac{V_2^2}{R_2} \\
& =\frac{\left(\frac{12}{5}\right)^2}{1}=\frac{144}{25} \left[R_2=1 \Omega \text { (given) }\right]\\
& =5.76 \mathrm{~W}
\end{aligned}
\end{aligned}
\)

Hint

(c) Given, inductance, \(L=20 \mathrm{mH}\)
\(
=20 \times 10^{-3} \mathrm{H}
\)
Capacitance, \(C=100 \mu \mathrm{~F}=100 \times 10^{-6} \mathrm{~F}\)
Resistance, \(R=50 \Omega \mathrm{emf}, V=10 \sin 314 t \dots(i)\)
\(\because\) The general equation of emf is given as \(V=V_0 \sin \omega t \dots(ii)\)
On comparing Eqs. (i) and (ii), we get
\(
\begin{aligned}
V_0 & =10 \mathrm{~V} \\
\omega & =314 \mathrm{rad} \mathrm{~s}^{-1}
\end{aligned}
\)
The power loss associated with the given AC circuit is given as
\(
\begin{aligned}
P & =V_{\mathrm{rms}} I_{\mathrm{rms}} \cos \phi \\
& =V_{\mathrm{rms}}\left(\frac{V_{\mathrm{rms}}}{Z}\right)\left(\frac{R}{Z}\right) \\
& =\left(\frac{V_{\mathrm{rms}}}{Z}\right)^2 R=\left(\frac{V_0}{\sqrt{2} \cdot Z}\right)^2 R \dots(iii)
\end{aligned}
\)
Impedance,
\(
\begin{aligned}
Z & =\sqrt{R^2+\left(X_L-X_C\right)^2} \\
& =\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}
\end{aligned}
\)
Substituting the given values in the above equation, we get
\(
\begin{aligned}
Z & =\sqrt{(50)^2+\left[\left(314 \times 20 \times 10^{-3}\right)-\frac{1}{314 \times 10^{-4}}\right]^2} \\
& =\sqrt{2500+\left[6280 \times 10^{-3}-0.00318 \times 10^4\right]^2} \\
& =\sqrt{2500+(-25.56)^2} \\
& =56.15 \Omega \simeq 56 \Omega
\end{aligned}
\)
Now, substituting this values in Eq. (iii), we get
\(
P=\left(\frac{10}{\sqrt{2} \times 56}\right)^2 \times 50=\frac{100}{2 \times 3136} \times 50=0.79 \mathrm{~W}
\)
Thus, the power loss in the circuit is 0.79 W.

Hint

(c) In an \(L-C-R\) circuit,
\(
\tan \phi=\frac{\omega L-\frac{1}{\omega C}}{R}
\)
\(\phi\) being the angle by which the current leads the voltage.
Given, \(\phi=45^{\circ}\)
\(
\begin{aligned}
\tan 45^{\circ} & =\frac{\omega L-\frac{1}{\omega C}}{R} \\
1 & =\frac{\omega L-\frac{1}{\omega C}}{R} \\
R & =\omega L-\frac{1}{\omega C} \\
\omega C & =\frac{1}{\omega L-R} \\
C & =\frac{1}{\omega(\omega L-R)} \\
& =\frac{1}{2 \pi f(2 \pi f L-R)}
\end{aligned}
\)

Hint

(a) We know that, for series \(R\)-C circuit
\(
V^2=V_C^2+V_R^2
\)

\(
\begin{array}{lr}
& 100=64+V_R^2 \\
\Rightarrow & V_R^2=36 \\
\Rightarrow & V_R=\sqrt{36}=6 \mathrm{~V} \\
\text { Also, } & \tan \phi=\frac{V_C}{V_R} \Rightarrow \tan \phi=\frac{8}{6} \\
& \tan \phi=\frac{4}{3} \\
\therefore & \phi=\tan ^{-1}\left(\frac{4}{3}\right)
\end{array}
\)

Hint

(d) Rise of current in \(L-R\) circuit is given by
\(
I=I_0\left(1-e^{-t / \tau}\right)
\)
where, \(\quad I_0=\frac{E}{R}=\frac{5}{5}=1 \mathrm{~A}\)
Now, \(\quad \tau=\frac{L}{R}=\frac{10}{5}=2 \mathrm{~s}\)
After 2 s , i.e. at \(t=2 \mathrm{~s}\)
Rise of current, \(I=\left(1-e^{-1}\right) \mathrm{A}\)

Hint

(b) For circuit \(A\),
Impedance, \(Z_A=\sqrt{R^2+\frac{1}{\omega^2 C^2}}\)
Current in circuit, \(I_R^A=\frac{V}{\sqrt{R^2+\left(1 / \omega^2 C^2\right)}} \dots(i)\)
Potential difference across \(C\),
\(
V_C^A=I_R^A \times \frac{1}{\omega C}=\frac{V}{\sqrt{(R \omega C)^2+1}}
\)
For circuit \(B\),
Impedance, \(Z_B=\sqrt{R^2+\frac{1}{(4 \omega)^2 C^2}}\)
Current in circuit, \(I_R^B=\frac{V}{\sqrt{R^2+\left[1 /(4 \omega C)^2\right]}}\)
Potential difference across \(C\),
\(
V_C^B=I_R^B \times \frac{1}{4 \omega C}=\frac{V}{\sqrt{(4 R \omega C)^2+1}} \dots(ii)
\)
Hence, from above Eqs. (i) and (ii), we conclude that as numerator of \(I_R^B\) is increased from \(I_R^A\) by a factor of 2.
So, \(I_R^B>I_R^A\) and \(V_C^A>V_C^B\).

Hint

(c) The resonant frequency is given by,
\(
f_0=\frac{1}{2 \pi} \sqrt{\frac{1}{L C}-\frac{R^2}{L^2}}
\)
where, \(R=\) resistance and \(L=\) inductance
\(
\begin{aligned}
\Rightarrow f_0 & =\frac{1}{2 \pi} \sqrt{\frac{1}{L C}-\frac{R^2}{L^2}} \\
& =\frac{1}{2 \times 3.14} \sqrt{\frac{1}{250 \times 10^{-12} \times 0.16 \times 10^{-3}}-\frac{20 \times 20}{\left(0.16 \times 10^{-3}\right)^2}} \\
& =8 \times 10^5 \mathrm{~Hz}
\end{aligned}
\)

Hint

(a) The resistance of inductor (coil), \(R=\frac{12}{24}=0.5 \Omega\)
If phase difference between current and emf is zero, then reactance of circuit must be zero.
The impedance of the circuit equals to resistance, i.e.
\(
Z=0.5 \Omega
\)
Now, rms current, \(I_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{Z}=\frac{24}{0.5}=48 \mathrm{~A}\)

Hint

(b) Given, output power \(=100 \mathrm{~W}\)
\(
V=220 \mathrm{~V}, I=0.5 \mathrm{~A}
\)
We know that, efficiency \(=\frac{\text { Output power }}{\text { Input power }} \times 100 \%\)
\(
=\frac{100}{220 \times 0.5} \times 100=90 \%
\)

Hint

(b) Given,
\(
\begin{aligned}
& V=100 \sqrt{2} \sin 100 t \mathrm{~V} \\
& C=1 \mu \mathrm{~F}
\end{aligned}
\)
The peak voltage, \(V_0=100 \sqrt{2} \mathrm{~V}\) and \(\omega=100 \mathrm{rad} / \mathrm{s}\)
\(
\begin{aligned}
&\therefore \mathrm{rms} \text { voltage, } V_{\mathrm{rms}}=\frac{V_0}{\sqrt{2}}=\frac{100 \sqrt{2}}{\sqrt{2}}=100 \mathrm{~V}\\
&\therefore \text { Current reading of the ammeter, }\\
&\begin{aligned}
I & =\frac{V_{\mathrm{rms}}}{X_C}=\frac{V_{\mathrm{rms}}}{\left(\frac{1}{\omega C}\right)}=\frac{100 \mathrm{~V}}{\frac{1}{100 \times 1 \times 10^{-6}}} \\
& =100 \times 100 \times 10^{-6}=10^{-2} \mathrm{~A} \\
& =10 \times 10^{-3} \mathrm{~A}=10 \mathrm{~mA}
\end{aligned}
\end{aligned}
\)

Hint

(a) The current through the AC circuit is given by \(i=3 \sin \omega t+4 \cos \omega t\)
The maximum value of current is
\(
I_{\max }=\sqrt{3^2+4^2}=\sqrt{25}=5 \mathrm{~A}
\)
The rms value of current, \(I_{\mathrm{rms}}=\frac{I_{\max }}{\sqrt{2}}=\frac{5}{\sqrt{2}} \mathrm{~A}\)

Hint

(c) We know that,
In \(L-C-R\) circuit, \(f=\frac{1}{2 \pi} \sqrt{\frac{1}{L C}}\)
For the same resonant frequency, the inductance should be changed from \(L\) to \(\frac{L}{4}\).

Hint

(c) In the inductor circuit, phase difference \((\phi)=90^{\circ}\) or \(\frac{\pi}{2}\)
We know that, \(P_{\mathrm{av}}=V_{\mathrm{rms}} I_{\mathrm{rms}} \cos \phi\)
\(
=V_{\mathrm{rms}} I_{\mathrm{rms}} \cos \frac{\pi}{2}
\)
\(
\therefore \quad P_{\text {av }}=0
\)

Hint

(c) Transformer is used to obtain desired AC voltage and current because a transformer is a device based on the principle of mutual induction, which is used for converting large AC at low voltage into small current at high voltage.

Hint

(b) As we know that, for transformer, \(\frac{V_1}{V_2}=\frac{I_2}{I_1}=\frac{N_1}{N_2}\) where all the symbols have their usual meanings. (Suffix 1 is for primary and 2 is for the secondary)
\(
\Rightarrow \quad \frac{N_1}{N_2}=\frac{1}{25}=\frac{2}{I}
\)
where, \(I=\) primary side current.
\(
\therefore \quad I=50 \mathrm{~A}
\)

Hint

(b) We know that,
\(
\begin{aligned}
& & V_S \times N_P & =V_P \times N_S \\
\Rightarrow & & \frac{N_S}{N_P} & =\frac{V_S}{V_P}
\end{aligned}
\)
where, \(\quad V_S=\) voltage across secondary coil
\(V_P=\) voltage across primary coil \(=220 \mathrm{~V}\)
\(N_P=\) number of turns is primary coil \(=1000\)
and \(\quad N_S=\) number of turns in secondary coil \(=50\)
\(
\begin{array}{ll}
\Rightarrow & \frac{50}{1000}=\frac{V_S}{220} \\
\Rightarrow & V_S=\frac{50 \times 220}{1000}=11 \mathrm{~V}
\end{array}
\)
Power in the secondary coil = Power in the primary coil
\(
\begin{aligned}
& & V_S I_S & =V_P I_P \\
\Rightarrow & & 11 \times I_S & =220 \times 1 \\
\Rightarrow & & I_S & =\frac{220}{11}=20 \mathrm{~A}
\end{aligned}
\)

Hint

(d) Average power,
\(\begin{aligned} P_{\mathrm{av}} & =E_{\mathrm{rms}} I_{\mathrm{rms}} \cos \phi \\ & =\frac{E_0}{\sqrt{2}} \cdot \frac{I_0}{\sqrt{2}} \cos \phi \\ & =\frac{100}{\sqrt{2}} \times \frac{100 \times 10^{-3}}{\sqrt{2}} \cos \frac{\pi}{3} \\ & =\frac{100 \times 100}{2} \times 10^{-3} \times \frac{1}{2}=2.5 \mathrm{~W}\end{aligned}\)

Hint

(a) As average power, \(\bar{P}=2 \mathrm{~W}, I_{\mathrm{rms}}=2 \mathrm{~A}\), impedance, \(Z=1 \Omega\)
Power factor \((\cos \phi)=\) ?
We know that, \(\bar{P}=V_{\mathrm{rms}} \times I_{\mathrm{rms}} \cdot \cos \phi\)
\(
\begin{aligned}
\bar{P} & =I_{\mathrm{rms}}^2 Z \cdot \cos \phi \quad\left[\because V_{\mathrm{rms}}=I_{\mathrm{rms}} \cdot Z\right] \\
\cos \phi & =\frac{\bar{P}}{I_{\mathrm{rms}}^2 \cdot Z}=\frac{2}{(2)^2 \times 1}=\frac{2}{4}=\frac{1}{2}=0.5
\end{aligned}
\)

Hint

(b) Given, \(V_L=60 \mathrm{~V}, V_C=30 \mathrm{~V}, V_R=40 \mathrm{~V}\)
In phasor diagram, \(V_C\) and \(V_L\) are in anti-phase to each other due to their \(90^{\circ}\) leading and lagging relationship with the circuit current \(I_S\).

\(
\begin{aligned}
V_S & =\sqrt{\left(V_L-V_C\right)^2+V_R^2}=\sqrt{(60-30)^2+(40)^2} \\
& =\sqrt{(30)^2+(40)^2} \\
& =\sqrt{900+1600} \\
& =\sqrt{2500}=50 \mathrm{~V}
\end{aligned}
\)