Past NEET Entrance Papers

Hint

(a) In an ideal transformer, the voltage ratio is related to the turns ratio by the formula:
\(
\frac{V_s}{V_p}=\frac{N_s}{N_p}
\)
Where:
\(V_s\) is the secondary voltage,
\(V_p\) is the primary voltage,
\(N_s\) is the number of turns in the secondary winding,
\(N_p\) is the number of turns in the primary winding.
Given that the turns ratio is \(\frac{N_p}{N_s}=\frac{1}{2}\), this implies:
\(
\frac{N_s}{N_p}=2
\)
Now, using the relationship for the voltage ratio:
\(
\frac{V_s}{V_p}=\frac{N_s}{N_p}=2
\)
Therefore, the voltage ratio \(V_s: V_p\) is \(2: 1\).

Hint

(a) The capacitive reactance \(X_C\) for a capacitor is given by the formula:
\(
X_C=\frac{1}{2 \pi f C}
\)
Where:
\(f\) is the frequency of the AC source \((50 \mathrm{~Hz})\),
\(C\) is the capacitance \(\left(10 \mu \mathrm{~F}=10 \times 10^{-6} \mathrm{~F}\right)\),
\(\pi \approx 3.14\).
Substituting the values:
\(
X_C=\frac{1}{2 \pi \times 50 \times 10 \times 10^{-6}}=\frac{1000}{3.14}
\)
\(
\begin{aligned}
& V_{\mathrm{rms}}=210 \mathrm{~V} \\
& i_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{X_C}=\frac{210}{X_C}
\end{aligned}
\)
\(
\begin{aligned}
& \text { Peak current }=\sqrt{2} i_{\mathrm{rms}}=\sqrt{2} \times \frac{210}{1000} \times 3.14=0.932 \\
& \quad=0.93 \mathrm{~A}
\end{aligned}
\)

Hint

(a) To find the factor with the same dimensions as the inverse of the resonance frequency, let’s first recall the formula for the resonance frequency \(f_0\) of an LC circuit:
\(
f_0=\frac{1}{2 \pi \sqrt{L C}}
\)
The inverse of the resonance frequency \(\frac{1}{f_0}\) would be:
\(
\frac{1}{f_0}=2 \pi \sqrt{L C}
\)
Now, let’s analyze the dimensional consistency:
The resonance frequency \(f_0\) has the dimension of time (specifically, \(T^{-1}\) ).
Therefore, the inverse resonance frequency \(\frac{1}{f_0}\) has the dimension of time (i.e., \(T\) ).
We need to find which of the given factors has the dimension of time \((T)\).
The dimensions of \(L\) (inductance) are \([L]=\mathrm{H}=\mathrm{kg} \cdot \mathrm{m}^2 \cdot \mathrm{~s}^{-2} \cdot \mathrm{~A}^{-2}\).
The dimensions of \(C\) (capacitance) are \([C]=\mathrm{F}=\mathrm{A}^2 \cdot \mathrm{~s}^4 \cdot \mathrm{~kg}^{-1} \cdot \mathrm{~m}^{-2}\).
The dimensions of \(L C\) will be \([L] \cdot[C]=\mathrm{s}^2\), so \(\sqrt{L C}\) has the dimension of time \(T\).
Thus, \(\sqrt{L C}\), has the same dimensions as the inverse of the resonance frequency.

Hint

(d) 

\(
\begin{aligned}
& V_L \text { leads current } I \text { by } \frac{\pi}{2} \\
& \therefore V_L=V_0 \sin \left(\omega t+\frac{\pi}{2}\right) \quad\left(\because I=I_0 \sin \omega t\right) \\
& V_0=I_0 X_L \\
& \Rightarrow V_L=I_0 X_L \cos (\omega t)=I_0 \omega L \cos (\omega t)
\end{aligned}
\)

Hint

(a) To calculate the current in the secondary circuit of the step-up transformer, we can use the principle of power conservation (assuming no power loss). The power input to the transformer is equal to the power output. Therefore, we can write:
\(
P_p=P_s
\)
Where:
\(P_p\) is the power input (primary side),
\(P_s\) is the power output (secondary side).
The power is given by:
\(
P=V \cdot I
\)
Thus, the input power \(P_p=V_p \cdot I_p\) and the output power \(P_s=V_s \cdot I_s\), where:
\(V_p=220 \mathrm{~V}\) is the primary voltage,
\(V_s=11000 \mathrm{~V}\) is the secondary voltage,
\(I_p\) is the primary current,
\(I_s\) is the secondary current.
Given that the transformer operates at 88 W on the secondary side:
\(
P_s=88 \mathrm{~W}
\)
Thus:
\(
88=11000 \cdot I_s
\)
Solving for \(I_s\) :
\(
I_s=\frac{88}{11000}
\)
Let’s calculate this:
It seems like I can’t do more advanced data analysis right now. Please try again later. However, I can guide you through the calculation.
You can calculate the secondary current \(I_s\) using the formula:
\(
I_s=\frac{P_s}{V_s}=\frac{88 \mathrm{~W}}{11000 \mathrm{~V}}=0.008 \mathrm{~A}=8 \mathrm{~mA}
\)

Hint

(b) The given equation for the charge in the circuit is:
\(
Q(t)=Q_0 e^{-\frac{Rt}{2 L}}
\)
Where:
\(Q_0\) is the initial charge at \(t=0\),
\(R\) is the resistance ( \(1.5 \Omega\) ),
\(L\) is the inductance \(\left(12 \mathrm{mH}=12 \times 10^{-3} \mathrm{H}\right)\),
\(t\) is the time,
\(Q(t)\) is the charge at time \(t\).
We are asked to find the time when the charge amplitude decreases to \(0.50 Q_0\), which means:
\(
Q(t)=0.50 Q_0
\)
Substituting this into the equation for \(Q(t)\) :
\(
0.50 Q_0=Q_0 e^{-\frac{Rt}{2 L}}
\)
Now, take the natural logarithm of both sides:
\(
\ln (0.50)=-\frac{R t}{2 L}
\)
Since \(\ln (0.50)=-\ln (2)\) and \(\ln (2) \approx 0.693\), we get:
\(
-\ln (2)=-\frac{R t}{2 L}
\)
Simplifying:
\(
\ln (2)=\frac{R t}{2 L}
\)
Substitute the given values for \(R\) and \(L\) :
\(
0.693=\frac{1.5 t}{2 \times 12 \times 10^{-3}}
\)
\(
t=0.011088 \text { seconds }=11.09 \mathrm{~ms}
\)

Hint

When an AC source is connected to a capacitor \(C\), the behavior of the capacitor in the circuit depends on the frequency of the AC signal. The capacitive reactance \(X_C\), which opposes the current flow in the circuit, is given by:
\(
X_C=\frac{1}{2 \pi f C}
\)
Where:
\(X_C\) is the capacitive reactance,
\(f\) is the frequency of the AC source,
\(C\) is the capacitance of the capacitor.

Effect of Decrease in Frequency:
When the operating frequency \(f\) decreases, the capacitive reactance \(X_C\) increases because \(X_C\) is inversely proportional to the frequency. As a result, the opposition to current flow in the circuit becomes greater.

Current Decreases: Since the current in an AC circuit with a capacitor is related to the voltage and the capacitive reactance by Ohm’s law \(I=\frac{V}{X_C}\), a higher \(X_C\) (due to lower frequency) will cause the current to decrease.

In summary, if the frequency decreases, the capacitive reactance increases, leading to a decrease in current.

Hint

(b) To calculate the current in the primary winding of the transformer, we can use the principle of power conservation, assuming the transformer is ideal (no losses).
Given:
The lamp operates at 12 V and 60 W , so the power output \(P_s\) on the secondary side is 60 W .
The primary voltage \(V_p=220 \mathrm{~V}\).
The secondary voltage \(V_s=12 \mathrm{~V}\).
Since the transformer is ideal, the power input \(P_p\) equals the power output \(P_s\) :
\(
P_p=P_s
\)
We can express the power in terms of voltage and current:
\(
P=V \times I
\)
So, the current in the secondary winding \(I_s\) is:
\(
I_s=\frac{P_s}{V_s}=\frac{60}{12}=5 \mathrm{~A}
\)
Now, we use the transformer equation for an ideal transformer, which relates the primary and secondary voltages and currents:
\(
\frac{V_p}{V_s}=\frac{I_s}{I_p}
\)
Rearranging to solve for the primary current \(I_p\) :
\(
I_p=\frac{V_s \cdot I_s}{V_p}
\)
Substituting the known values:
\(
I_p=\frac{12 \times 5}{220}=\frac{60}{220} \approx 0.27 \mathrm{~A}
\)
Thus, the current in the primary winding is approximately: 0.27 A

Hint

(a) The goal is to find the frequency at which resonance occurs in a series LCR circuit. To achieve resonance, the inductive reactance \(\left(X_L\right)\) should equal the capacitive reactance \(\left(X_C\right)\).
Given are:
Inductance, \(L=10 \mathrm{mH}=10 \times 10^{-3} \mathrm{H}\)
Capacitance, \(C=1 \mu \mathrm{~F}=1 \times 10^{-6} \mathrm{~F}\)
Resistance, \(R=100 \Omega\) (which does not directly impact the resonance frequency)
Resonance occurs when the angular frequency \((\omega)\) meets the condition \(\omega L=\frac{1}{\omega C}\), leading to the formula for calculating the resonance frequency \((f)\) as
\(
f=\frac{\omega}{2 \pi}=\frac{1}{2 \pi \sqrt{L C}}
\)
Substituting the given values:
\(
f=\frac{1}{2 \pi \sqrt{10 \times 10^{-3} \cdot 1 \times 10^{-6}}}=1.59 \mathrm{kHz}
\)
Thus, the frequency at which resonance occurs in this LCR circuit is 1.59 kHz.

Hint

(d)

\(
\begin{aligned}
& X_L=2 \pi \mathrm{fL} \\
& =2 \pi \times 50 \times \frac{50}{\pi} \times 10^{-3} \\
& X_L=5 \Omega
\end{aligned}
\)
\(
\begin{aligned}
& \mathrm{X}_{\mathrm{C}}=\frac{1}{2 \pi \mathrm{fC}}=\frac{10^6}{2 \pi \times 50 \times \frac{10^3}{\pi}} \\
& \mathrm{X}_{\mathrm{C}}=10 \Omega \\
& \mathrm{Z}=\sqrt{(10)^2+(10-5)^2} \\
& \mathrm{Z}=\sqrt{125}=5 \sqrt{5} \Omega
\end{aligned}
\)

Hint

(d) In a \(LR\) circuit, phase difference between current and voltage, is given by tan \(L\)
\(
\begin{aligned}
& \tan \phi=\frac{\mathrm{X}_{\mathrm{L}}}{\mathrm{R}}=\frac{L \omega}{R} \\
& \tan \phi=\frac{1}{\pi} \frac{100 \pi}{100}=1
\end{aligned}
\)
\(\phi=45^{\circ}\) voltage is leading the current.

Hint

(a)

\(
\begin{aligned}
& Z_1=\sqrt{X_L^2+R^2} \quad X_L=0 \text { (D.C. circuit) } \\
& Z_1=10 \Omega \\
& Z_2=\sqrt{X_C^2+R^2} \\
& X_C=\frac{1}{2 \pi \times 50 \times \frac{10^3}{\pi} \times 10^{-6}}=10 \Omega \\
& Z_2=\sqrt{(10)^2+(10)^2} \\
& =10 \sqrt{2} \\
& Z_1<Z_2
\end{aligned}
\)

Hint

(a) The maximum power is dissipated for an AC in a purely resistive circuit.
Explanation: In a purely resistive circuit, the voltage and current are in phase, meaning the power factor is 1 , which leads to the maximum power dissipation.

Power consumed in an AC circut is given by the formula
\(
\mathrm{P}=\mathrm{E}_{\mathrm{rms}} \cdot \mathrm{I}_{\mathrm{rms}} \cos \phi
\)
For a purely resistive circuit \(\cos \phi=1\) So, power consumption will be maximum.

Hint

(c) For very high frequency, \(\omega L\) will be very high, so branch will not conduct current (behaves as open circuit) which has inductance; whereas, at high frequency, \(\frac{1}{\omega \mathrm{C}}\) will be very small, so effective reactance of the capacitor will be zero (behaves as short). So, from the given circuit, total impedance will be \(1+2=3 \Omega\)

Hint

(c) The formula for calculating the magnetic energy stored in an inductor is given by:
\(
E=\frac{1}{2} L I^2
\)
where:
\(E\) is the energy stored (in joules, J ),
\(L\) is the inductance of the inductor (in henrys, H ),
\(I\) is the current flowing through the inductor (in amperes, A).
Plugging the given values into the formula:
\(
\begin{aligned}
& E=\frac{1}{2} \times 4 \mu H \times(2 A)^2 \\
& E=\frac{1}{2} \times 4 \times 10^{-6} H \times 4 A^2 \\
& E=2 \times 10^{-6} H \cdot 4 \\
& E=8 \times 10^{-6} J
\end{aligned}
\)
Hence, the energy stored in the inductor is \(8 \mu J\).

Hint

(c) We know,
RMS value of A.C. \(E_{r m s}=\frac{E_0}{\sqrt{2}}\)
\(
E_0=\sqrt{2} E_{r m s}
\)

Hint

(c) At resonance,
\(
\begin{aligned}
& \omega L=\frac{1}{\omega C} \\
& \omega=\frac{1}{\sqrt{L C}}=\frac{1}{\sqrt{10 \times 10 \times 10^{-6}}} \\
& \omega=100 \\
& \omega=2 \pi f \Rightarrow f=\frac{\omega}{2 \pi} \\
& \nu_0=f_0=\frac{100}{2 \pi}=\frac{50}{\pi} \mathrm{~Hz}, \omega=100 \\
& \nu=f=\frac{100}{2 \pi}=\frac{50}{\pi}
\end{aligned}
\)

Hint

(a) To find the instantaneous power supplied to a pure inductor, we need to use the voltage and current expressions. For ar inductor, the voltage and current are out of phase by 90 degrees. Given the voltage \(V=V_m \sin (\omega t)\), the current throug the inductor is \(I=I_m \sin \left(\omega t-\frac{\pi}{2}\right)=I_m \cos (\omega t)\). The instantaneous power \(P(t)\) is given by \(P(t)=V(t) \cdot I(t)\). Substituting the expressions for \(V(t)\) and \(I(t)\). we get \(P(t)=V_m \sin (\omega t) \cdot I_m \cos (\omega t)\). Using the trigonometric identity \(\sin (2 \theta)=\) \(2 \sin (\theta) \cos (\theta)\). we can \(\operatorname{simplify}\) this to \(P(t)=\frac{V_m I_m}{2} \sin (2 \omega t)\).

Step by Step Solution:
Step 1: Write the expressions for voltage and current in the inductor: \(V=V_m \sin (\omega t)\) and \(I=I_m \cos (\omega t)\).

Step 2: Express the instantaneous power \(P(t)\) as the product of voltage and current: \(P(t)=V(t) \cdot I(t)\).

Step 3: Substitute the expressions for \(V(t)\) and \(I(t): P(t)=V_m \sin (\omega t) \cdot I_m \cos (\omega t)\).

Step 4: Use the trigonometric identity \(\sin (2 \theta)=2 \sin (\theta) \cos (\theta)\) to simplify: \(P(t)=\frac{V_m I_m}{2} \sin (2 \omega t)\).

Hint

(d)

\(
\begin{aligned}
&\begin{aligned}
& V=V_0 \sin \omega t \\
& V=\frac{q}{C} \\
& q=C V \sin \omega t \\
& \frac{d q}{d t}=C V \frac{d}{d t} \sin \omega t \\
& i=C V \omega \cos \omega t \\
& i=C V \omega \sin \left(\omega t+\frac{\pi}{2}\right)
\end{aligned}\\
&\text { In a.c circuit current lead by voltage by a phase } \frac{\pi}{2}
\end{aligned}
\)
Thus statement I is correct while II is incorrect.

Hint

(c) We know, for A.C. source
\(
\begin{aligned}
& X_L=\omega L \\
& =2 \pi f(L) \\
& =100 \pi\left(2 \times 10^{-3}\right) \\
& =0.2 \pi \Omega=0.628 \Omega
\end{aligned}
\)
For D.C. source, the inductor behaves as a closed circuit offering no resistance at all (at steady state) as \(\omega=0\) (For D.C.)
\(
\therefore X_L=0 \Omega
\)

Hint

(a)

\(
\begin{aligned}
&P_{a v g}=V_{r m s} \times I_{r m s} \times \cos \phi\\
&\text { Here, } \phi=0^{\circ} \text { as current and voltage are in same phase in a resistor. }\\
&\Rightarrow 100=200 \times I_{r m s} \Rightarrow I_{r m s}=\frac{1}{2} \Rightarrow \frac{I_0}{\sqrt{2}}=\frac{1}{2} \Rightarrow I_0=\frac{1}{\sqrt{2}}=0.707 \mathrm{~A}
\end{aligned}
\)

Hint

(b) 

\(
\begin{aligned}
&{I}_0=10 \sqrt{2} \mathrm{~A}\\
&I_{\mathrm{RMS}}=\frac{I_0}{\sqrt{2}}=10 \mathrm{~A}
\end{aligned}
\)
\(
\begin{aligned}
& V_{R M S}=\sqrt{V_R^2+\left(V_L-V_C\right)^2} \\
& =\sqrt{(40)^2+(40-10)^2} \\
& =50 \mathrm{~V} \\
& \text { Impedance, } Z=\frac{V_{R M S}}{I_{R M}}=\frac{50 V}{10 \mathrm{~V}}=5 \Omega
\end{aligned}
\)

Hint

(c) For an ideal transformer (no power losses), the input power equals the output power: \(\boldsymbol{P}_{\text {in }}=\boldsymbol{P}_{\text {out }}\).
Power can be calculated using the formula: \(P=V I\), where \(V\) is voltage and \(I\) is current.
Calculate the current in the primary circuit using the power equation and the principle of power conservation in an ideal transformer.

Step 1: Calculate the current in the secondary circuit

Use the formula \(P=V_s I_s\) to find \(I_s\).
Rearrange the formula to solve for \(I_s: I_s=\frac{P}{V_s}\).
Substitute the given values: \(I_s=\frac{44 \mathrm{~W}}{11 \mathrm{~V}}\).
Calculate: \(\boldsymbol{I}_s=4 \mathrm{~A}\).

Step 2: Calculate the current in the primary circuit

Since power is conserved in an ideal transformer, \(\boldsymbol{P}_{\text {in }}=\boldsymbol{P}_{\text {out }}\).
Therefore, \(V_p I_p=V_s I_s\).
Rearrange the formula to solve for \(I_p: I_p=\frac{V_s I_s}{V_p}\).
Substitute the known values: \(I_p=\frac{11 \mathrm{~V} \cdot 4 \mathrm{~A}}{220 \mathrm{~V}}\).
Calculate: \(I_p=\frac{44}{220} \mathrm{~A}=0.2 \mathrm{~A}\).

Hint

(a)

\(
\begin{aligned}
& Q=\frac{\omega}{\Delta \omega}=\frac{\omega L}{R} \Rightarrow \Delta \omega=R / L=\frac{50}{4}=8 \mathrm{rad} / \mathrm{sec} \\
& \omega_0=\frac{1}{\sqrt{L C}}=\frac{1}{\sqrt{5 \times 80 \times 10^{-6}}}=50 \mathrm{rad} / \mathrm{sec} \\
& \omega_{\min }=\omega_0-\frac{\Delta \omega}{2}=46 \mathrm{rad} / \mathrm{sec} \\
& \omega_{\max }=\omega_0-\frac{\Delta \omega}{2}=54 \mathrm{rad} / \mathrm{sec}
\end{aligned}
\)

Hint

(b) Given, \(\mathrm{V}=\mathrm{V}_0 \sin \omega \mathrm{t}\)
We know, \(q=C V\)
Now displacement current \(I_d\) is given by,
\(
\begin{aligned}
& \mathrm{I}_{\mathrm{d}}=\frac{d q}{d t}=\frac{C d V}{d t} \\
& I_d=C\left(V_0 \omega \cos \omega t\right) \\
& =V_0 \omega C \cos \omega t
\end{aligned}
\)

Hint

(b)

\(
\begin{aligned}
&\mathrm{i}_{\mathrm{rms}}=\mathrm{C} \omega \varepsilon_{\mathrm{rms}}\\
&\mathrm{C}=40 \times 10^{-6} \mathrm{~F}\\
&\omega=2 \pi f=100 \pi\\
&\varepsilon_{\mathrm{rms}}=200 \mathrm{~V}\\
&\begin{aligned}
& \mathrm{i}_{\mathrm{rms}}=200 \times 40 \times 10^{-6} \times 2 \pi \times 50 \\
& =2.5 \mathrm{~A}
\end{aligned}
\end{aligned}
\)

Hint

(b) Find the relationship between \(\boldsymbol{X}_{\boldsymbol{L}}, \boldsymbol{X}_{\boldsymbol{C}}\) and \(\boldsymbol{R}\), then calculate the power factor.
Step 1: Find the relationship between \(X_C\) and \(R\)
When \(L\) is removed, \(\tan \phi=\frac{X_C}{R}\).
Given \(\phi=\frac{\pi}{3}\), so \(\tan \frac{\pi}{3}=\frac{X_C}{R}\).
\(
\begin{aligned}
& \tan \frac{\pi}{3}=\sqrt{3}, \text { so } \sqrt{3}=\frac{X_C}{R} . \\
& X_C=R \sqrt{3} .
\end{aligned}
\)
Step 2: Find the relationship between \(X_L\) and \(R\)
When \(C\) is removed, \(\tan \phi=\frac{X_L}{R}\).
Given \(\phi=\frac{\pi}{3}\), so \(\tan \frac{\pi}{3}=\frac{X_L}{R}\).
\(
\begin{aligned}
& \tan \frac{\pi}{3}=\sqrt{3}, \text { so } \sqrt{3}=\frac{X_L}{R} . \\
& X_L=R \sqrt{3} .
\end{aligned}
\)
Step 3: Find the phase difference in the LCR circuit
In the LCR circuit, \(\tan \phi=\frac{X_L-X_C}{R}\).
Substitute \(X_L\) and \(X_C: \tan \phi=\frac{R \sqrt{3}-R \sqrt{3}}{R}\).
\(
\begin{aligned}
& \tan \phi=0 . \\
& \phi=0 .
\end{aligned}
\)
Step 4: Calculate the power factor
Power factor is \(\cos \phi\).
\(
\cos 0=1 .
\)
The power factor of the circuit is 1.

Hint

(a)

\(
L=\frac{\mu_0 N^2\left(\pi r^2\right)}{l}
\)
After inserting the iron rod
\(
\begin{array}{ll}
L_{\text {final }}=\frac{\mu N^2\left(\pi r^2\right)}{l} & \mu>\mu_0 \\
e=-L\left(\frac{d i}{d t}\right) &
\end{array}
\)
As \(L\) increases, emf induced across the inductor will increase leaving less voltage across the bulb. Therefore, the glow of the light bulb decreases.

Hint

(b) For Direct current magnitude and direction remain the same with time. For Alternating current both magnitude and direction change with time.
\(\therefore\) All 4 cases satisfy the condition of AC current.

Hint

(a) In a circuit with an AC source, RMS current is given as:
\(I_{\mathrm{rms}}=\mathrm{V}_{\mathrm{rms}} / \mathrm{Z} \dots(1)\)
where, \(\mathrm{V}_{\text {rms }}=\) potential of AC source, \(\mathrm{Z}=\) impedance
Also, \(\mathrm{Z}=\sqrt{R^2+\left(X_L-X_C\right)^2} \dots(2)\)
In a circuit with a DC source, the current is defined as:
\(
\mathrm{I}=\mathrm{V} / \mathrm{R} \dots(3)
\)
Given:
\(
\mathrm{V}_{\mathrm{ms}}=12 \mathrm{~V}, \mathrm{I}_{\mathrm{rms}}=0.2 \mathrm{~A}, \mathrm{~V}=12 \mathrm{~V}, \mathrm{I}=0.4 \mathrm{~A}
\)
From equation (1) we get,
\(
Z=12 / 0.2=60 \Omega
\)
And from equation (3) we get,
\(
R=12 / 0.4=30 \Omega
\)
\(\because\) There is current in the circuit at a steady state (DC source) hence capacitor cannot be in series with Resistance.
Also, since \(Z>R\) and from equation (3) we can conclude there must be another component in the series which is nothing but Inductor.
\(\therefore\) It is a series LR circuit.

Hint

(a) Impedance \(Z\) in an ac circuit is
\(Z=\sqrt{R^2+\left(X_C-X_L\right)^2}\); where \(X_C=\) capacitive reactance and \(X_L=\) inductive reactance.
Also \(X_C=\frac{1}{\omega C}\) and \(X_L=\omega L\)
\(
\therefore Z=\sqrt{(50)^2+\left(\frac{1}{314 \times 100 \times 10^{-6}}-314 \times 20 \times 10^{-3}\right)^2}
\)
or \(Z=56 \Omega\)
The power loss in the circuit is \(P_{a v}=\left(\frac{V_{r m s}}{Z}\right)^2 R\)
\(
\therefore \quad P_{a v}=\left(\frac{10}{(\sqrt{2}) 56}\right)^2 \times 50=0.79 \mathrm{~W}
\)

Hint

(b)

At time, \(t=0\) i.e., when switch is closed, inductor in the circuit provides very high resistance (open circuit) while capacitor starts charging with maximum current (low resistance).
Equivalent circuit of the given circuit.

Current drawn from battery,
\(
i=\frac{\varepsilon}{(R / 2)}=\frac{2 \varepsilon}{R}=\frac{2 \times 18}{9}=4 \mathrm{~A}
\)

Hint

(c) Here, \(V_R=80 \mathrm{~V}, V_C=40 \mathrm{~V}, V_L=100 \mathrm{~V}\)
Power factor, \(\cos \phi=\frac{R}{Z}=\frac{V_R}{V}=\frac{V_R}{\sqrt{V_R^2+\left(V_L-V_C\right)^2}}\)
\(=\frac{80}{\sqrt{(80)^2+(100-40)^2}}=\frac{80}{100}=0.8\)

Hint

(a) Peak voltage: \(V_0=V_{r m s} \sqrt{2}\)
Impedance in a series RC circuit: \(\boldsymbol{Z}=\sqrt{R^2+X_C^2}\)
Peak current: \(I_0=\frac{V_0}{Z}\)
The displacement current is equal to the conduction current in a series RC circuit.
\(
\begin{aligned}
& Z=\sqrt{R^2+X_C^2} \\
& Z=\sqrt{(100 \Omega)^2+(100 \Omega)^2} \\
& Z=\sqrt{10000 \Omega^2+10000 \Omega^2} \\
& Z=\sqrt{20000 \Omega^2} \\
& Z=100 \sqrt{2} \Omega
\end{aligned}
\)
Use the formula for peak voltage
\(V_0=V_{r m s} \sqrt{2}\)
\(V_0=220 \mathrm{~V} \cdot \sqrt{2}\)
\(V_0=220 \sqrt{2} V\)
Use the formula for peak current:
\(I_0=\frac{V_0}{Z}\)
\(I_0=\frac{220 \sqrt{2} V}{100 \sqrt{2} \Omega}\)
\(I_0=\frac{220}{100} A\)
\(I_0=2.2 \mathrm{~A}\)
The peak displacement current is equal to the peak current:
\(I_{d 0}=I_0\)
\(I_{d 0}=2.2 \mathrm{~A}\)
The peak value of the displacement current is 2.2 A.

Hint

(a) When an ideal capacitor is connected with an ac voltage source, current leads voltage by \(90^{\circ}\). Since, energy stored in capacitor during charging is spent in maintaining charge on the capacitor during discharging. Hence over a full cycle the capacitor does not consume any energy from the voltage source.

Power, \(\mathrm{P}=\mathrm{V}_{\mathrm{rms}} \mathrm{I}_{\mathrm{rms}} \cos \phi\)
\(\Rightarrow \cos \phi=0\). (as \(\phi=90^{\circ}\) between \(\mathrm{I}_{\mathrm{rms}}\) and \(\mathrm{V}_{\mathrm{rms}}\) across a capacitor)

Therefore, power consumed \(=0\) in one complete cycle across a capacitor.

Hint

(c) Quality factor of an \(L-C-R\) circuit is given by,
\(
\begin{aligned}
& Q=\frac{1}{R} \sqrt{\frac{L}{C}} \\
& Q_1=\frac{1}{20} \sqrt{\frac{1.5}{35 \times 10^{-6}}}=50 \times \sqrt{\frac{3}{70}}=10.35 \\
& Q_2=\frac{1}{25} \times \sqrt{\frac{2.5}{45 \times 10^{-6}}}=40 \times \sqrt{\frac{5}{90}}=9.43 \\
& Q_3=\frac{1}{15} \sqrt{\frac{3.5}{30 \times 10^{-6}}}=\frac{100}{15} \sqrt{\frac{35}{3}}=22.77 \\
& Q_4=\frac{1}{25} \times \sqrt{\frac{1.5}{45 \times 10^{-6}}}=\frac{40}{\sqrt{30}}=7.30
\end{aligned}
\)
Clearly \(Q_3\) is maximum of \(Q_1, Q_2, Q_3\), and \(Q_4\). Hence, option (c) should be selected for better tuning of an \(L-C-R\) circuit.

Hint

\(
\begin{aligned}
&\begin{aligned}
& \text { (d) Given: } \mathrm{L}=20 \mathrm{mH} ; \mathrm{C}=50 \mu \mathrm{~F} ; \mathrm{R}=40 \Omega \\
& \mathrm{~V}=10 \sin 340 \mathrm{t} \\
& \therefore \quad \mathrm{~V}_{\text {runs }}=\frac{10}{\sqrt{2}} \\
& \mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}=\frac{1}{340 \times 50 \times 10^{-6}}=58.8 \Omega \\
& \mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}=340 \times 20 \times 10^{-3}=6.8 \Omega \\
& \text { Impedance, } \mathrm{Z}=\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{C}}-\mathrm{X}_{\mathrm{L}}\right)^2} \\
& =\sqrt{40^2+(58.8-6.8)^2}=\sqrt{4304} \Omega
\end{aligned}\\
&\text { Power loss in A.C. circuit, }\\
&\begin{aligned}
& \mathrm{P}=\mathrm{i}_{\mathrm{ms}}^2 \mathrm{R}=\left(\frac{\mathrm{V}_{\mathrm{rms}}}{\mathrm{Z}}\right)^2 \mathrm{R} \\
& =\left(\frac{10 / \sqrt{2}}{\sqrt{4304}}\right)^2 \times 40=\frac{50 \times 40}{4304} \simeq 0.51 \mathrm{~W}
\end{aligned}
\end{aligned}
\)

Hint

(b)

Current through resistor, \(i=\) Current in the circuit
\(
=\frac{V_0}{\sqrt{R^2+X_C^2}}=\frac{V_0}{\sqrt{R^2+(1 / \omega C)^2}}
\)
Voltage across capacitor, \(V=i X_C\)
\(
=\frac{V_0}{\sqrt{R^2+(1 / \omega C)^2}} \times \frac{1}{\omega C}=\frac{V_0}{\sqrt{R^2 \omega^2 C^2+1}}
\)
As \(C_a<C_b\) \(\therefore \quad i_a<i_b\) and \(V_a>V_b\)

Hint

(a) For pure resistor circuit, power
\(
\begin{aligned}
& \mathrm{P}=\frac{V_{r m s}^2}{R} \\
& \Rightarrow V_{r m s}^2=\mathrm{PR}
\end{aligned}
\)
For L-R series circuit, power
\(
\begin{aligned}
& \mathrm{P}^{\prime}={V}_{\mathrm{rms}} {I}_{\mathrm{rms}} \cos \phi \\
& =V_{r m s} \times \frac{V_{r m s}}{Z} \times \frac{R}{Z} \\
& =\frac{V_{r m s}^2 R}{Z^2}=\frac{P R \times R}{Z^2} \\
& \therefore \mathrm{P}^{\prime}=P\left(\frac{R}{Z}\right)^2
\end{aligned}
\)

Hint

(b) Here, Efficiency of the transformer, \(\eta=90 \%\)

Input power, \(P_{\text {in }}=3 \mathrm{~kW}=3 \times 10^3 \mathrm{~W}=3000 \mathrm{~W}\)
Voltage across the primary coil, \(V_p=200 \mathrm{~V}\)
Current in the secondary coil, \(I_s=6 \mathrm{~A}\)
As \(P_{\text {in }}=I_p V_p\)
\(\therefore \quad\) Current in the primary coil,
\(
I_p=\frac{P_{\mathrm{in}}}{V_p}=\frac{3000 \mathrm{~W}}{200 \mathrm{~V}}=15 \mathrm{~A}
\)
Efficiency of the transformer,
\(
\begin{aligned}
& \eta=\frac{P_{\text {out }}}{P_{\text {in }}}=\frac{V_s I_s}{V_p I_p} \\
\therefore \quad & \frac{90}{100}=\frac{6 V_s}{3000} \text { or } V_s=\frac{90 \times 3000}{100 \times 6}=450 \mathrm{~V}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (c): Transformer cannot work on dc. }\\
&\therefore \quad V_s=0 \text { and } I_s=0
\end{aligned}
\)

Hint

(b) By inserting iron rod in the coil,
\(\mathrm{L} \uparrow \mathrm{Z} \uparrow \mathrm{I} \downarrow\) so brightness \(\downarrow\)

Note: Inductive reactance, \(X_L=\omega L\). On inserting iron rod in the coil, \(L\) increases. As a result Impedance \(\mathrm{Z}=\sqrt{X_L^2+R^2}=\sqrt{(\omega L)^2+R^2}\) also increases.
So, Current, \(I=\frac{V}{Z}\) decreases.

Hint

(d) When \(L\) is removed, the phase difference between the voltage and current is
\(
\begin{aligned}
& \tan \phi_1=\frac{X_C}{R} \\
& \tan \frac{\pi}{3}=\frac{X_C}{R} \text { or } X_C=R \tan 60^{\circ} \text { or } X_C=\sqrt{3} R
\end{aligned}
\)
When \(C\) is removed, the phase difference between the voltage and current is \(\tan \phi_2=\frac{X_L}{R}\) or \(\tan \frac{\pi}{3}=\frac{X_L}{R}\) or \(X_L=R \tan 60^{\circ}=\sqrt{3} R\)
As \(X_L=X_C\), the series \(L C R\) circuit is in resonance. Impedance of the circuit,
\(
Z=\sqrt{R^2+\left(X_L-X_C\right)^2}=R
\)
Power factor, \(\cos \phi=\frac{R}{Z}=\frac{R}{R}=1\)

Hint

(d) Given : \(i=\frac{1}{\sqrt{2}} \sin (100 \pi t)\) ampere Compare it with \(i=i_0 \sin (\omega t)\), we get
\(
i_0=\frac{1}{\sqrt{2}} \mathrm{~A}
\)
Given : \(e=\frac{1}{\sqrt{2}} \sin \left(100 \pi t+\frac{\pi}{3}\right)\) volt
Compare it with \(e=e_0 \sin (\omega t+\phi)\), we get
\(
\begin{aligned}
& e_0=\frac{1}{\sqrt{2}} \mathrm{~V}, \phi=\frac{\pi}{3} \\
\therefore \quad & i_{\mathrm{rms}}=\frac{i_0}{\sqrt{2}}=\frac{1}{2} \mathrm{~A} \text { and } e_{\mathrm{rms}}=\frac{e_0}{\sqrt{2}}=\frac{1}{2} \mathrm{~V}
\end{aligned}
\)
Average power consumed in the circuit,
\(
\begin{aligned}
& P=i_{\mathrm{rms}} e_{\mathrm{rms}} \cos \phi \\
& =\left(\frac{1}{2}\right)\left(\frac{1}{2}\right) \cos \frac{\pi}{3}=\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)=\frac{1}{8} \mathrm{~W}
\end{aligned}
\)

Hint

(d) \(V=-L \frac{d i}{d t}\)

Here \(\frac{d i}{d t}\) is + ve for \(\frac{T}{2}\) time and \(\frac{d i}{d t}\) is – ve for next \(\frac{T}{2}\) time.

Hint

(c) The given equation of alternating voltage is
\(
e=200 \sqrt{2} \sin 100 t \dots(i)
\)
The standard equation of alternating voltage is
\(
e=e_0 \sin \omega t \dots(ii)
\)
Comparing (i) and (ii), we get
\(
e_0=200 \sqrt{2} \mathrm{~V}, \omega=100 \mathrm{rad} \mathrm{~s}^{-1}
\)
The capacitive reactance is
\(
X_C=\frac{1}{\omega C}=\frac{1}{100 \times 1 \times 10^{-6}} \Omega
\)
The r.m.s. value of the current in the circuit is
\(
\begin{aligned}
& i_{\text {r.m.s. }}=\frac{v_{\text {r.m.s. }}}{X_C}=\frac{e_0 / \sqrt{2}}{1 / \omega C}=\frac{(200 \sqrt{2} / \sqrt{2})}{\left(1 / 100 \times 10^{-6}\right)} \\
& =200 \times 100 \times 10^{-6} \mathrm{~A}=2 \times 10^{-2} \mathrm{~A}=20 \mathrm{~mA}
\end{aligned}
\)

Hint

(a) Here, \(R=3 \Omega, X_L=3 \Omega\)
The phase difference between the applied voltage and the current in the circuit is
\(
\tan \phi=\frac{X_L}{R}=\frac{3 \Omega}{3 \Omega}=1 \quad \text { or } \quad \phi=\tan ^{-1}(1)=\frac{\pi}{4}
\)

Hint

(b) 

\(
\begin{aligned}
& V=V_0 \text { for } 0 \leq t \leq \frac{T}{2} \\
& V=0 \text { for } \frac{T}{2} \leq t \leq T
\end{aligned}
\)
\(
V_{\mathrm{rms}}=\left[\frac{\int_0^T V^2 d t}{\int_0^T d t}\right]^{1 / 2}=\left[\frac{\int_0^{T / 2} V_0^2 d t+\int_{T / 2}^T(0) d t}{\int_0^T d t}\right]^{1 / 2}
\)
\(
\begin{aligned}
& =\left[\frac{V_0^2}{T}[t]_0^{T / 2}\right]^{1 / 2}=\left[\frac{V_0^2}{T}\left(\frac{T}{2}\right)\right]^{1 / 2}=\left[\frac{V_0^2}{2}\right]^{1 / 2} \\
& \therefore V_{\mathrm{rms}}=\frac{V_0}{\sqrt{2}}
\end{aligned}
\)

Alternate:

\(
\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{(\mathrm{T} / 2) \mathrm{V}_0^2+0}{\mathrm{~T}}}=\frac{\mathrm{V}_0}{\sqrt{2}} .
\)

Hint

(b)

\(
\begin{aligned}
&\text { Current flowing in the coil is }\\
&\mathrm{I}=\frac{200}{Z}=\frac{200}{\sqrt{R^2+(\omega L)^2}}=\frac{200}{\sqrt{(30)^2+(40)^2}}=4 \mathrm{~A}
\end{aligned}
\)

Hint

(b) As \(V_L=V_C=300 \mathrm{~V}\), therefore the given series \(L C R\) circuit is in resonance.
\(
\therefore \quad V_R=V=220 \mathrm{~V}, Z=R=100 \Omega
\)
Current, \(I=\frac{V}{Z}=\frac{220 \mathrm{~V}}{100 \Omega}=2.2 \mathrm{~A}\)
Hence, the reading of the voltmeter \(V_3\) is 220 V and the reading of ammeter \(A\) is 2.2 A .

Hint

(d) Here, Input voltage, \(V_p=220 \mathrm{~V}\)
Output voltage, \(V_s=440 \mathrm{~V}\)
Input current, \(I_p=\) ?
Output current, \(I_s=2 \mathrm{~A}\)
Efficiency of the transformer, \(\eta=80 \%\)
Efficiency of the transformer, \(\eta=\frac{\text { Output power }}{\text { Input power }}\)
\(
\begin{aligned}
\eta=\frac{V_s I_s}{V_p I_p} \text { or } I_p & =\frac{V_s I_s}{\eta V_p}=\frac{(440 \mathrm{~V})(2 \mathrm{~A})}{\left(\frac{80}{100}\right)(220 \mathrm{~V})} \\
& =\frac{(440 \mathrm{~V})(2 \mathrm{~A})(100)}{(80)(220 \mathrm{~V})}=5 \mathrm{~A}
\end{aligned}
\)

Hint

(d) In case of oscillatory discharge of a capacitor through an inductor, charge at instant \(t\) is given by
\(
\begin{gathered}
\quad q=q_0 \cos \omega t ; \text { where, } \omega=\frac{1}{\sqrt{L C}} \\
\therefore \cos \omega t=\frac{q}{q_0}=\frac{C V_2}{C V_1}=\frac{V_2}{V_1} \quad(\because q=C V)
\end{gathered}
\)
\(
\begin{aligned}
&\text { Current through the inductor }\\
&\begin{aligned}
& I=\frac{d q}{d t}=\frac{d}{d t}\left(q_0 \cos \omega t\right)=-q_0 \omega \sin \omega t \\
& |I|=C V_1 \frac{1}{\sqrt{L C}}\left[1-\cos ^2 \omega t\right]^{1 / 2} \\
& \\
& =V_1 \sqrt{\frac{C}{L}}\left[1-\left(\frac{V_2}{V_1}\right)^2\right]^{1 / 2}=\left[\frac{C\left(V_1^2-V_2^2\right)}{L}\right]^{1 / 2}
\end{aligned}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (d) : Average power, } P=E_{\text {r.m. } \mathrm{s}} I_{\mathrm{r} . \mathrm{m} . \mathrm{s}} \cos \phi\\
&\begin{aligned}
& Z=\sqrt{R^2+\left(X_L-X_C\right)^2}, \cos \phi=\frac{R}{Z} \\
& \text { But } I_{\text {r.m.s }}=\frac{E_{\text {r.m.s }}}{Z} \quad \therefore \quad P=E_{\text {r.m.s }}^2 \cdot \frac{R}{Z^2} \\
& \therefore P=E_{\text {r.m.s }}^2 \frac{R}{\left\{R^2+\left(X_L-X_C\right)^2\right\}}=\frac{\varepsilon^2 R}{\left[R^2+\left(L \omega-\frac{1}{C \omega}\right)^2\right]}
\end{aligned}
\end{aligned}
\)

Hint

(a) \(\text { Average power }=\frac{E_0 I_0}{2} \cos \phi\)

Hint

(d) In series \(L C R\), current is maximum at resonance.
\(\therefore\) Resonant frequency, \(\omega=\frac{1}{\sqrt{L C}}\)
\(\therefore \quad \omega^2=\frac{1}{L C} \quad\) or, \(L=\frac{1}{\omega^2 C}\)
Given \(\omega=1000 \mathrm{~s}^{-1}\) and \(C=10 \mu \mathrm{~F}\)
\(
\therefore L=\frac{1}{1000 \times 1000 \times 10 \times 10^{-6}}=0.1 \mathrm{H}=100 \mathrm{mH}
\)

Hint

(c) Given : Output power \(P=100 \mathrm{~W}\)
Voltage across primary \(V_p=220 \mathrm{~V}\)
Current in the primary \(I_p=0.5 \mathrm{~A}\)
Efficiency of a transformer
\(
\begin{aligned}
\eta & =\frac{\text { output power }}{\text { input power }} \times 100 \\
& =\frac{P}{V_p I_p} \times 100=\frac{100}{220 \times 0.5} \times 100=90 \%
\end{aligned}
\)

Hint

(a) No. of turns across primary \(N_p=50\)
Number of turns across secondary \(N_s=1500\)
Magnetic flux linked with primary, \(\phi=\phi_0+4 t\)
\(\therefore \quad\) Voltage across the primary,
\(
\begin{aligned}
& V_p=\frac{d \phi}{d t}=\frac{d}{d t}\left(\phi_0+4 t\right)=4 \mathrm{volt} \\
& \frac{V_s}{V_p}=\frac{N_s}{N_p} \text { or } V_s=\left(\frac{1500}{50}\right) \times 4=120 \mathrm{~V}
\end{aligned}
\)

Hint

(a) The core of a transformer is laminated to minimise the energy losses due to eddy currents.

Hint

(c) Impedance of series LCR is
\(
\begin{aligned}
& Z=\sqrt{R^2+\left(X_L-X_C\right)^2} \\
& =\sqrt{8^2+(31-25)^2} \\
& =\sqrt{64+36}=10 \Omega
\end{aligned}
\)
Power factor, \(\cos \phi=\frac{R}{Z}=\frac{8}{10}=0.8\)

Hint

\(
\begin{aligned}
&\text { (d) : Frequency of } L C \text { oscillation }=\frac{1}{2 \pi \sqrt{L C}}\\
&\begin{aligned}
& \text { or, } \frac{f_1}{f_2}=\left(\frac{L_2 C_2}{L_1 C_1}\right)^{1 / 2}=\left(\frac{2 L \times 4 C}{L \times C}\right)^{1 / 2}=(8)^{1 / 2} \\
& \therefore \frac{f_1}{f_2}=2 \sqrt{2} \Rightarrow f_2=\frac{f_1}{2 \sqrt{2}} \text { or, } f_2=\frac{f}{2 \sqrt{2}} \quad\left(\because f_1=f\right)
\end{aligned}
\end{aligned}
\)

Hint

(d) \(\tan \phi=\frac{X_C-X_L}{R}\)
or \(\tan \left(\frac{\pi}{4}\right)=\frac{\frac{1}{\omega C}-\omega L}{R}\)
\(R=\frac{1}{\omega C}-\omega L\)
or \((R+2 \pi f L)=\frac{1}{2 \pi f C}\)
or \(C=\frac{1}{2 \pi f(R+2 \pi f L)}\)

Hint

\(
\begin{aligned}
&\text { (a) Time constant of } L R \text { circuit is } \tau=L / R \text {. }\\
&\tau=\frac{40}{8}=5 \mathrm{~s}
\end{aligned}
\)

Hint

(c) The impedance \(Z\) of a series \(L C R\) circuit is given by, \(Z=\sqrt{R^2+\left(X_L-X_C\right)^2}\)
where \(X_L=\omega L\) and \(X_C=\frac{1}{\omega C}, \omega\) is angular frequency.
At resonance, \(X_L=X_C\), hence \(Z=R\).
\(\therefore \quad V_R=V\) (supply voltage)
\(\therefore\) r.m.s. current, \(I=\frac{V_R}{R}=\frac{V}{R}\)
\(\therefore \quad\) Power loss \(=I^2 R=V^2 / R\)

Hint

\(
\begin{aligned}
&\text { (c) Capacitive reactance, } X=\frac{1}{\omega C}=\frac{1}{2 \pi f C}\\
&\begin{aligned}
& \Rightarrow X \propto \frac{1}{f C} \\
& \therefore \frac{X^{\prime}}{X}=\frac{f}{f^{\prime}} \times \frac{C}{C^{\prime}}=\frac{f}{2 f} \times \frac{C}{2 C}=\frac{1}{4} \\
& \Rightarrow X^{\prime}=\frac{X}{4}
\end{aligned}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (d) Quality factor }\\
&\begin{aligned}
& =\frac{\text { Potential drop across capacitor or inductor }}{\text { Potential drop across } \mathrm{R}} \\
& =\frac{I \omega L}{I R}=\frac{\omega L}{R}
\end{aligned}
\end{aligned}
\)

Hint

(d) At \(t=0\)
(i) capacitor offers negligible resistance (acts as short).
(ii) inductor offers large resistance (acts as open) to current flow.

Hint

(b) \(\frac{E_P}{E_S}=\frac{I_S}{I_P}=\frac{N_P}{N_S}=\frac{1}{25}\);
Here, \(I_S=2 \mathrm{~A}\)
\(
I_P=25 I_S=50 \mathrm{~A}
\)

Hint

(c) Turns on primary winding \(=500\); Turns on secondary winding \(=5000\); Primary winding voltage \(\left(E_p\right)\) \(=20 \mathrm{~V}\) and frequency \(=50 \mathrm{~Hz}\).
\(
\begin{aligned}
& \quad \frac{N_s}{N_p}=\frac{E_s}{E_p} \text { or } \frac{5000}{500}=\frac{E_s}{20} \\
& \text { or } E_s=\frac{5000 \times 20}{500}=200 \mathrm{~V}
\end{aligned}
\)
and frequency remains the same. Therefore secondary winding will have an output of \(200 \mathrm{~V}, 50 \mathrm{~Hz}\).

Hint

(a) Power dissipated \(=V_{\mathrm{rms}} \cdot I_{\mathrm{rms}}\)
\(
=\left(V_{\mathrm{rms}}\right)\left(I_{\mathrm{rms}}\right) \cos \theta
\)
Hence, power dissipated depends upon phase difference.

Hint

(a) Given : Supply voltage \(\left(V_{a c}\right)=200 \mathrm{~V}\)
Inductive reactance \(\left(X_L\right)=50 \Omega\)
Capacitive reactance \(\left(X_C\right)=50 \Omega\)
Ohmic resistance \((R)=10 \Omega\).
We know that impedance of the LCR circuit \((Z)\)
\(
\begin{aligned}
& =\sqrt{\left\{\left(X_L-X_C\right)^2+R^2\right\}} \\
& =\sqrt{\left\{(50-50)^2+(10)^2\right\}}=10 \Omega
\end{aligned}
\)

Hint

(b) Initially, the current will pass through the capacitor (and not through the resistance which is parallel to the capacitor). So effective resistance in the circuit is \(R_{A B}\). Therefore the current in the resistor is 2 mA . After some time, the capacitor will become fully charged and will be in its steady state. Now no current will pass through the capacitor and the effective resistance of the circuit that is
\(
(1000+1000)=2000 \Omega
\)
Therefore final current in the resistor
\(
=\frac{V}{R}=\frac{2}{2000}=1 \times 10^{-3} \mathrm{~A}=1 \mathrm{~mA}
\)

Hint

(b) Current \((I)=5 \sin (100 t-\pi / 2)\) and voltage \((V)\) \(=200 \sin (100 t)\). Comparing the given equation, with the standard equation, we find that phase between current and voltage is \(\phi=\frac{\pi}{2}=90^{\circ}\)
Power consumption \(P=I_{\mathrm{rms}} V_{\mathrm{rms}} \cos \phi=I_{\mathrm{rms}} V_{\mathrm{rms}} \cos 90^{\circ}=0\)

Hint

(b) For resonance condition, the impedance will be minimum and the current will be maximum. This is only possible when \(X_L=X_C\).
Therefore \(\tan \phi=\frac{X_L+X_C}{R}=0\) or \(\phi=0\).

Hint

\(
\text { (d) } \quad I_{\mathrm{rms}}=\frac{I_0}{\sqrt{2}}
\)

Hint

(a) In an A.C. circuit with inductance coil, the voltage \(V\) leads the current \(I\) by a phase difference of \(90^{\circ}\) or the current \(I\) lags behind the voltage \(V\) by a phase difference of \(90^{\circ}\). Thus the voltage goes on decreasing with the increase in time as shown in the graph (a).

Hint

(a) The major portion of the A.C. flows on the surface of the wire. So where a thick wire is required, a number of thin wires are joined together to give an equivalent effect of a thick wire. Therefore multiple strands are suitable for transporting A.C. Similarly multiple strands can also be used for D.C.

Explanation:

step 1: Compare the resistance of the two cables.
The total cross-sectional area of the second cable is \(10 \times \frac{A}{10}=A\).
Since both cables have the same length and total cross-sectional area, their resistances are equal for DC.
\(
\begin{aligned}
& R_1=\rho \frac{l}{A} \\
& R_2=\rho \frac{l}{10 \cdot \frac{A}{10}}=\rho \frac{l}{A}
\end{aligned}
\)
Therefore, \(\boldsymbol{R}_1=\boldsymbol{R}_2\).

Step 2: Analyze the skin effect for AC.
For AC, the current tends to flow on the surface of the conductor (skin effect).
The multi-wire cable has a larger surface area compared to the single-wire cable.
Therefore, the multi-wire cable is more suitable for AC due to reduced skin effect losses.

Step 3: Conclude the suitability for AC and DC .
Multi-wire cable is better for AC due to the skin effect.
Both cables are equally suitable for DC as they have the same resistance.

Therefore, Multi-wire cable is more suitable for AC, while both cables are equally suitable for DC.

Hint

(c) The time constant for \(R-C\). circuit, \(\tau=C R\)
Growth of charge in a circuit containing capacitance and resistance is given by the formula,
\(
q=q_0\left(1-e^{-t / C R}\right)
\)
\(C R\) is known as time constant in this formula.

Hint

(c) \(I_0=\frac{E_0}{R}=\frac{n B A \omega}{R}\)
Given, \(n=1, B=10^{-2} \mathrm{~T}\),
\(
\begin{aligned}
& A=\pi(0.3)^2 \mathrm{~m}^2, R=\pi^2 \\
& f=(200 / 60) \text { and } \omega=2 \pi(200 / 60)
\end{aligned}
\)
Substituting these values and solving, we get
\(
I_0=6 \times 10^{-3} \mathrm{~A}=6 \mathrm{~mA}
\)

Hint

(a) Eddy currents are produced when a metal is kept in a varying magnetic field.