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If the rms current in a 50 Hz ac circuit is 5 A , the value of the current \(1 / 300\) seconds after its value becomes zero is
(b) \(I=I_o \sin \omega t \ldots(1)\)
We know the relation of peak current and rms current is \(I_o=\sqrt{2} I_{r m s}\), so we will use the value of rms current for the calculation of peak current
Substitute the values in the relation, therefore we get
\(
\begin{aligned}
& I_o=\sqrt{2} \times 5 \mathrm{~A} \\
& \mathrm{I}_o=5 \sqrt{2} \mathrm{~A}
\end{aligned}
\)
Now we have the value of \(\omega\) and \(I_o\), by substituting these values in equation (1) we can calculate the current afterl/300 seconds, so
\(\)
\begin{aligned}
& I=I_o \sin \omega t \\
& I=(5 \sqrt{2} \mathrm{~A}) \sin \left(100 \pi \mathrm{rad} / \mathrm{s} \times \frac{1}{300} \mathrm{~s}\right) \\
& I=(5 \sqrt{2} \mathrm{~A}) \sin \left(\frac{\pi}{3}\right) \\
& I=(5 \sqrt{2} \mathrm{~A}) \times \frac{\sqrt{3}}{2}
\end{aligned}
\(\)
On further solving, we get
\(\)
I=5 \sqrt{\frac{3}{2}} \mathrm{~A}
\(\)
An alternating current generator has an internal resistance Rg and an internal reactance Xg . It is used to supply power to a passive load consisting of a resistance \(R_g\) and a reactance \(X_L\). For maximum power to be delivered from the generator to the load, the value of \(X_L\) is equal to
(c) For maximum power to be delivered from the generator (or internal reactance \(X_g\) ) to the load (of reactance, \(x_L\) ),
\(
\begin{aligned}
& =>X_L+X_g=0 \text { (the total reactance must vanish) } \\
& \Rightarrow>X_L=-X_g
\end{aligned}
\)
When a voltage measuring device is connected to AC mains, the meter shows the steady input voltage of 220 V . This means
(c) The voltmeter connected to AC mains reads mean value \(\left(\left\langle v^2\right\rangle\right)\) and is calibrated in such a way that it gives value of \(\left\langle v^2\right\rangle\), which is multiplied by \(\sqrt{2}\) to get \(\mathrm{V}_{\text {rms. }}\)
To reduce the reasonant frequency in an LCR series circuit with a generator
(b) Resonant frequency in an \(L-C-R\) circuit is given by
\(
f_0=\frac{1}{2 \pi \sqrt{L C}}
\)
If \(L\) or \(C\) increases, the resonant frequency will reduce.
To increase capacitance, we must connect another capacitor parallel to the first.
Which of the following combinations should be selected for better tuning of an \(LCR\) circuit used for communication?
(c) Quality factor \((Q)\) of an \(L-C-R\) circuit is given by,
\(
Q=\frac{1}{R} \sqrt{\frac{L}{C}}
\)
where \(R\) is resistance, \(L\) is inductance and \(C\) is capacitance of the circuit. To make \(Q\) high,
\(R\) should be low, \(L\) should be high and \(C\) should be low.
These conditions are best satisfied by the values given in option (c).
An inductor of reactance \(1 \Omega\) and a resistor of \(2 \Omega\) are connected in series to the terminals of a \(6 \mathrm{~V}(\mathrm{rms})\) a.c. source. The power dissipated in the circuit is
(c) According to the problem, \(\mathrm{X_1710:29 PM 2/7/2025L}=1 \Omega, \mathrm{R}=2 \Omega\), \(E_{\mathrm{rms}}=6 \mathrm{~V}, P_{\mathrm{av}}=?\)
\(
\begin{aligned}
&\text { Average power dissipated in the circuit }\\
&P_{\mathrm{av}}=E_{\mathrm{rms}} I_{\mathrm{rms}} \cos \phi \dots(i)
\end{aligned}
\)
\(
\begin{aligned}
I_{\mathrm{rms}} & =\frac{E_{\mathrm{rms}}}{Z} \\
Z & =\sqrt{R^2+X_L^2} \\
& =\sqrt{4+1}=\sqrt{5} \\
I_{\mathrm{mms}} & =\frac{6}{\sqrt{5}} \mathrm{~A} \\
\cos \phi & =\frac{R}{Z}=\frac{2}{\sqrt{5}}
\end{aligned}
\)
\(P_{\mathrm{av}}=6 \times \frac{6}{\sqrt{5}} \times \frac{2}{\sqrt{5}} \text { [from Eq. (i)] }\)
\(
=\frac{72}{\sqrt{5} \sqrt{5}}=\frac{72}{5}=14.4 \mathrm{~W}
\)
The output of a step-down transformer is measured to be 24 V when connected to a 12 watt light bulb. The value of the peak current is
(a) According to the problem output/secondary voltage \(V_S=24 \mathrm{~V}\) Power associated with secondary \(P_{\mathrm{S}}=12 \mathrm{~W}\)
\(
I_S=\frac{P_S}{V_S}=\frac{12}{24}=0.5 \mathrm{~A}
\)
Amplitude of the current in the secondary winding
\(
\begin{aligned}
I_0 & =I_S \sqrt{2} \\
& =(0.5)(1.414)=0.707=\frac{1}{\sqrt{2}} \mathrm{~A}
\end{aligned}
\)
As the frequency of an ac circuit increases, the current first increases and then decreases. What combination of circuit elements is most likely to comprise the circuit?
(a, d) Compare the given circuit by predicting the variation in their reactances with frequency. So, that then we can decide the elements.
Reactance of an inductor of inductance \(L\) is \(X_L=2 \pi f L\), where \(f\) is the frequency of the \(A C\) circuit.
\(X_{\mathrm{c}}=\) Reactance of the capacitive circuit
\(
=\frac{1}{2 \pi f C}
\)
With an increase in frequency ( \(f\) ) of an AC circuit, \(R\) remains constant, inductive reactance \(\left(X_L\right)\) increases and capacitive reactance \(\left(X_c\right)\) decreases.
For an \(L-C-R\) circuit,
\(Z=\) Impedance of the circuit
\(
\begin{aligned}
& =\sqrt{R^2+\left(X_L-X_C\right)^2} \\
& =\sqrt{R^2+\left(2 \pi v L-\frac{1}{2 \pi v C}\right)^2}
\end{aligned}
\)
As frequency \((f)\) increases, \(Z\) decreases and at certain value of the frequency known as resonant frequency \(\left(f_0\right)\), impedance \(Z\) is minimum that is \(Z_{\min }=R\) current varies inversely with impedance and at \(Z_{\min }\) current is maximum.
In an alternating current circuit consisting of elements in series, the current increases on increasing the frequency of supply. Which of the following elements are likely to constitute the circuit?
(c, d) This is the similar problem as we discussed above. In this problem, the current increases on increasing the frequency of supply. Hence, the reactance of the circuit must be decreased as increase in frequency. So, one element that must be connected is capacitor. We can also connect a resistor in series.
For a capacitive circuit,
\(
X_C=1 / \omega C=1 / 2 \pi f C
\)
When frequency increases, \(\mathrm{X}_{\mathrm{C}}\) decreases. Hence current in the circuit increases.
Electrical energy is transmitted over large distances at high alternating voltages. Which of the following statements is (are) correct?
(a, b, d) As we know that Power \(=I_{\mathrm{rms}}^2 R\)
So to decrease power loss \(I_{r m s}\), and \(R\) must be lower for a constant power supply. To decrease \(I_{r m s}\), \(\mathrm{V}_{\text {rms }}\) must be increased by step up transformer to get same power in step up transformer.
For a given power level, we find that
\(P=E_{\mathrm{rms}} I_{\mathrm{rms}}\) ( \(I_{\mathrm{rms}}\) is low when \(E_{\mathrm{rms}}\) is high)
Power loss \(=I_{\text {rms }}^2 R=\) low \(\quad\left(\because I_{\text {rms }}\right.\) is low \()\)
Now at the receiving end high voltage is reduced by using step-down transformers.
For an \(L C R\) circuit, the power transferred from the driving source to the driven oscillator is \(P=I^2 Z \cos \phi\).
(a, b, c) According to question power transferred,
\(
P=I^2 Z \cos \phi
\)
where \(I\) is the current, \(Z=\) Impedance and \(\cos \phi\) is power factor
As power factor,
\(
\begin{aligned}
\cos \phi & =\frac{R}{Z} \\
R & >0 \text { and } Z>0 \\
\cos \phi & >0 \Rightarrow P>0
\end{aligned}
\)
When an AC voltage of 220 V is applied to the capacitor \(C\)
(c, d) When the AC voltage is applied to the capacitor, the plate connected to the positive terminal will be at higher potential and the plate connected to the negative terminal will be at lower potential.
The plate with positive charge will be at higher potential and the plate with negative charge will be at lower potential. So, we can say that the charge is in phase with the applied voltage.
Power applied to a circuit is \(\quad P_{\mathrm{av}}=V_{\mathrm{rms}} I_{\mathrm{rms}} \cos \phi\)
For capacitive circuit, \(\quad \phi=90^{\circ}\)
\(
\begin{aligned}
\cos \phi & =0 \\
P_{\mathrm{av}} & =\text { Power delivered }=0
\end{aligned}
\)
The line that draws power supply to your house from street has
(a, d) Alternating currents are used for household supplies, which are having zero average value over a cycle.
The line is having some resistance, so power factor \(\cos \phi=\frac{R}{Z} \neq 0\)
So, \(\quad \phi \neq \pi / 2 \Rightarrow \phi<\pi / 2\)
i.e., phase lies between 0 and \(\pi / 2\).
Important point: The average value of alternating quantity for one complete cycle is zero.
The average value of ac over half cycle ( \(t=0[latex] to [latex]T / 2[latex] )
[latex]
i_{\mathrm{av}}=\frac{\int_0^{T / 2} i d t}{\int_0^{T / 2} d t}=\frac{2 i_0}{\pi}=0.637 i_0=63.7 \% \text { of } i_0
\)
\(
\text { Similarly } V_{\mathrm{av}}=\frac{2 V_0}{\pi}=0.637 V_0=63.7 \% \text { of } V_0 \text {. }
\)
A capacitor acts as an infinite resistance for
(a) In DC Current, the frequency is zero, hence the reactance is maxium, if we go through the expression for reactance.
\(
X_C=\frac{1}{\omega C}=\frac{1}{2 \pi f C}
\)
As \(f=0\) for dc, \(X_c=\)infinite.
An AC source producing emf
\(
\varepsilon=\varepsilon_0\left[\cos \left(100 \pi \mathrm{~s}^{-1}\right) t+\cos \left(500 \pi \mathrm{~s}^{-1}\right) t\right]
\)
is connected in series with a capacitor and a resistor. The steady-state current in the circuit is found to be
\(
i=i_1 \cos \left[\left(100 \pi \mathrm{~s}^{-1}\right) t+\varphi_1\right]+i_2 \cos \left[\left(500 \pi \mathrm{~s}^{-1}\right) t+\phi_2\right]
\)
(c)
\(
i_1=\frac{{\epsilon_0}}{\sqrt{R^2+\left(\frac{1}{w_1 C}\right)^2}}=\frac{\epsilon_0}{Z_1},
\)
where \(\omega_1=100 \pi\)
\(
i_2=\frac{\epsilon_0}{\sqrt{R^2+\left(\frac{1}{w_2 C}\right)^2}}=\frac{\epsilon_0}{Z_2}
\)
where \(\omega_2=500 \pi\)
So, \(Z_1>Z_2\), therefore \(i_1<i_2\).
The peak voltage in a 220 V AC source is
(c) Given root mean square voltage,
\(
V_{r m s}=220 \mathrm{~V}
\)
Peak voltage will be given by,
\(
\begin{aligned}
V_{\text {peak }} & =\sqrt{2} V_{\text {rms }} \\
V_{\text {peak }} & =\sqrt{2}(220) \\
V_{\text {peak }} & \approx 310 \mathrm{~V}
\end{aligned}
\)
An AC source is rated \(220 \mathrm{~V}, 50 \mathrm{~Hz}\). The average voltage is calculated in a time interval of 0.01 s. It
(b)
\(
\begin{aligned}
& V_0=V_{rms} \times \sqrt{ } 2=220 \times \sqrt{ } 2=3 311 \\
& f=50 \mathrm{~Hz} \\
& \omega=2 \pi f=100 \pi \\
& T=\frac{1}{f}=\frac{1}{50}=\frac{2}{100}=0.02 \mathrm{sec}
\end{aligned}
\)
0.01 s is half of the time period which is T/2. Average value of a sine function over half time period will not be 0.
But if we take the sine function with a phase shift of \(\phi =\pi/2\)
\(
V=V_0 \sin (\omega t + \pi/2)
\)
The average value of this function over T/2 will be 0 as there is equal negative area and equal positive area in the sine function.
The magnetic field energy in an inductor changes from maximum value to minimum value in 5.0 ms when connected to an AC source. The frequency of the source is
(b) The magnetic field energy in an inductor is given by,
\(
E=\frac{1}{2} L i^2
\)
The magnetic energy will be maximum when the current will reach its peak value, \(i_0\), and it will be minimum when the current will become zero.
From the above graph of alternating current, we can see that current reduces from maximum to zero in \(T / 4\) time, where \(T\) is the time period.
So, in this case, \(T / 4=5 \mathrm{~ms}\)
\(
\Rightarrow \mathrm{T}=20 \mathrm{~ms}
\)
\(\therefore\) Frequency, \(V=\frac{1}{T}=\frac{1}{20 \times 10^{-3}}=50 \mathrm{~Hz}\)
Which of the following plots may represent the reactance of a series \(L C\) combination?
(d) The reactance of a series LC circuit is given by,
\(
\begin{aligned}
& X=X_L-X_c=\omega L-\frac{1}{\text { \omega C }} \\
& \Rightarrow X=2 \pi f L-\frac{1}{2 \pi f C}
\end{aligned}
[latex]
The correct relation between the reactance and [latex]f\) is represented by the graph in option (d). \(X=0\), when \(X_L=X_c\)
Thus plot d represent the reactance of a series LC combination.
A series AC circuit has a resistance of \(4 \Omega\) and a reactance of \(3 \Omega\). The impedance of the circuit is
(a)
\(
Z=\sqrt{R^2+X^2}
\)
\(
\begin{aligned}
& Z=\sqrt{(4 \Omega)^2+(3 \Omega)^2} \\
& Z=\sqrt{16 \Omega^2+9 \Omega^2} \\
& Z=\sqrt{25 \Omega^2} \\
& Z=5 \Omega
\end{aligned}
\)
Transformers are used
(b) Understanding the Function of Transformers:
Transformers operate on the principle of electromagnetic induction. They require a changing magnetic field to induce a voltage in the secondary coil.
Evaluating Option 1 (DC Circuit Only):
In a DC circuit, the current is constant and does not change over time. Since transformers rely on changing magnetic fields to induce electromotive force (EMF), a constant DC current will not produce a changing magnetic field. Therefore, transformers cannot operate in a DC circuit.
Evaluating Option 2 (AC Circuit Only):
In an AC circuit, the current alternates and changes direction periodically. This change in current creates a changing magnetic field, which induces EMF in the secondary coil of the transformer. Thus, transformers are designed to work specifically with alternating current (AC).
An alternating current is given by
\(
i=i_1 \cos \omega t+i_2 \sin \omega t
\)
The rms current is given by
(c) From given equation it is clear that \(i_1\) and \(i_2\) are perpendicular to each other
\(
i_0=\sqrt{i_1^2+i_2^2}
\)
\(
i_{r m s}=\frac{i_0}{\sqrt{2}}
\)
\(
=\sqrt{\frac{i_1^2+i_2^2}{2}}
\)
An alternating current having peak value 14 A is used to heat a metal wire. To produce the same heating effect, a constant current \(i\) can be used where \(i\) is
(d)
\(
I_{r m s}=\frac{I_0}{\sqrt{2}}=\frac{14 \mathrm{~A}}{\sqrt{2}} \approx 10 \mathrm{~A}
\)
A constant current of 2.8 A exists in a resistor. The rms current is
The constant current is equal to the rms value of current. So, \(I_{r m s}=2.8 \mathrm{~A}\)
An inductor, a resistor and a capacitor are joined in series with an AC source. As the frequency of the source is slightly increased from a very low value, the reactance
(a) An inductor, a resistor, and a capacitor are joined in series with an AC source. As the frequency of the source is slightly increased from a very low value, the reactance of the inductor increases.
The reactance of an inductor, \(X_L=\omega L\)
The reactance of inductor \(X_C=\frac{1}{\omega C}\)
The reactance of a circuit is zero. It is possible that the circuit contains
(a, d) A circuit can have zero reactance if it contains either no inductors or capacitors, or if it contains both an inductor and a capacitor where their individual reactances cancel each other out.
In an AC series circuit, the instantaneous current is zero when the instantaneous voltage is maximum. Connected to the source may be a
(a, b, d) In an AC circuit, when the instantaneous current is zero while the instantaneous voltage is maximum, this indicates that the current and voltage are out of phase by 90 degrees.
This phase relationship is characteristic of either a pure inductor or a pure capacitor.
For a pure inductor, the current lags the voltage by 90 degrees. For a pure capacitor, the current leads the voltage by 90 degrees. In both cases, the current and voltage are out of phase by 90 degrees, meaning when the voltage is at its maximum or minimum, the current is zero.
This is possible when phase difference is \(\frac{\pi}{2}\) between current and voltage.
An inductor-coil having some resistance is connected to an AC source. Which of the following quantities have zero average value over a cycle?
The quantities with zero average value over a cycle in an inductor-coil connected to an AC source are the (a) Current and (b) Induced emf in the inductor.
Explanation:
Current and Induced EMF: In an AC circuit, the current and induced EMF are sinusoidal functions, meaning they oscillate between positive and negative values. Over a complete cycle, the positive and negative portions cancel each other out, resulting in a zero average value.
Why other options are incorrect:
(c) Joule heat:
Joule heat is proportional to the square of the current. Since the current has a zero average value, the Joule heat generated over a cycle will also be positive, resulting in a non-zero average value. (\(H_{a v g}=i_{r m s}^2 R\)) which is non-zero.
(d) Magnetic energy stored in the inductor:
The magnetic energy stored in an inductor is proportional to the square of the current. Similar to Joule heat, the average magnetic energy stored will be positive over a cycle due to the current’s squared nature. \(U_{\text {avg }}=\frac{1}{2} { L i_{rms}^2 }, \text { which is also non zero }\).
The AC voltage across a resistance can be measured using
(b) The correct answer is (b) a hot-wire voltmeter.
Explanation:
Hot-wire voltmeters measure AC voltage by utilizing the heating effect of the current. As the current flows through a wire, it heats up, and the change in temperature (which is proportional to the square of the current) is used to indicate the voltage. This method works for both AC and DC, as it relies on the heating effect, which is independent of the direction of current flow.
To convert mechanical energy into electrical energy, one can use
(a, b) To convert mechanical energy into electrical energy, you would use a (a) DC dynamo or (b) AC dynamo.
Explanation:
Dynamo:
Both DC and AC dynamos (also known as electric generators) are specifically designed to convert mechanical energy into electrical energy using electromagnetic induction.
Motor:
An electric motor, on the other hand, does the opposite; it converts electrical energy into mechanical energy.
Transformer:
A transformer is used to change the voltage of electrical energy, not to convert mechanical energy into electrical energy.
An AC source rated 100 V (rms) supplies a current of 10 A (rms) to a circuit. The average power delivered by the source
(b, d)
\(
\begin{aligned}
&\begin{aligned}
& P_{\alpha v}=E_v I_v \cos \phi=100 \times 10 \cos \phi \\
& =1000 \cos \phi \text { Watt }
\end{aligned}\\
&\text { So depending upon value of } \phi, P_{a v} \text { can be equal to or less than } 1000 \mathrm{~W} \text {. }
\end{aligned}
\)
The frequency of a sinusoidal wave \(E=0.40 \cos [2000 t+0.80]\) would be
\(
\begin{aligned}
&\text { (d) } 2 \pi f=2000\\
&\therefore f=\frac{1000}{\pi} \mathrm{~Hz}
\end{aligned}
\)
A 220 V AC is more dangerous than 220 V DC because
(d) Here rms value of ac \(V_{r m s}=220 \mathrm{~V}\). But the peak value of the given ac supply, \(V_0=220 \sqrt{2}=311 \mathrm{~V}\), In the case of 220 V dc , the peak value is the same 220 V. That is why the ac is more dangerous than dc of the same voltage.
Alternating current is transmitted to take places
(a) Alternating current is transmitted at high voltage and low current. This is done to minimize power loss during transmission over long distances.
Explanation:
Power Loss:
Power loss in transmission lines is proportional to the square of the current ( \(\mathrm{P}=\mathrm{I}^2 \mathrm{R}\), where \(R\) is resistance).
High Voltage, Low Current:
By transmitting at high voltage and low current, the power loss due to resistance in the transmission lines is minimized.
Efficiency:
This method of transmission is more efficient because less energy is lost as heat.
Transformers:
Transformers can easily step up or step down AC voltage, making high-voltage transmission feasible and then safer for local use.
An AC voltage is given by \(E=E_0 \sin \frac{2 \pi t}{T}\)
Then, the mean value of voltage calculated over any time interval of \(T / 2\) second
(d)
\(
\begin{aligned}
& E_{\text {mean }}=\frac{1}{\frac{T}{2}-0} \int_0^{\frac{T}{2}} E_0 \sin \left(\frac{2 \pi t}{T}\right) d t \\
& E_{\text {mean }}=\frac{2}{T} E_0 \int_0^{\frac{T}{2}} \sin \left(\frac{2 \pi t}{T}\right) d t
\end{aligned}
\)
The integral of \(\sin (a x)\) is \(-\frac{1}{a} \cos (a x)\).
Here, \(a=\frac{2 \pi}{T}\).
\(
\begin{aligned}
& \int_0^{\frac{T}{2}} \sin \left(\frac{2 \pi t}{T}\right) d t=\left[-\frac{T}{2 \pi} \cos \left(\frac{2 \pi t}{T}\right)\right]_0^{\frac{T}{2}} \\
& =-\frac{T}{2 \pi}\left[\cos \left(\frac{2 \pi\left(\frac{T}{2}\right)}{T}\right)-\cos (0)\right] \\
& =-\frac{T}{2 \pi}[\cos (\pi)-\cos (0)] \\
& =-\frac{T}{2 \pi}[-1-1]=-\frac{T}{2 \pi}(-2)=\frac{T}{\pi}
\end{aligned}
\)
\(
E_{\text {mean }}=\frac{2}{T} E_0\left(\frac{T}{\pi}\right)=\frac{2 E_0}{\pi}
\)
If the interval is from \(t_1=\frac{T}{4}\) to \(t_2=\frac{3 T}{4}\), the integral would be:
\(
\begin{aligned}
& \int_{\frac{T}{4}}^{\frac{3 T}{4}} \sin \left(\frac{2 \pi t}{T}\right) d t=\left[-\frac{T}{2 \pi} \cos \left(\frac{2 \pi t}{T}\right)\right]_{\frac{T}{4}}^{\frac{3 T}{4}} \\
& =-\frac{T}{2 \pi}\left[\cos \left(\frac{3 \pi}{2}\right)-\cos \left(\frac{\pi}{2}\right)\right] \\
& =-\frac{T}{2 \pi}[0-0]=0
\end{aligned}
\)
In this case, the mean value is zero.
The mean value is \(\frac{2 E_0}{\pi}\) for the interval \(\left[0, \frac{T}{2}\right]\).
The mean value is 0 for the interval \(\left[\frac{T}{4}, \frac{3 T}{4}\right]\).
Therefore, the mean value calculated over any time interval of \(\frac{T}{2}\) seconds may be zero.
\(220 \mathrm{~V}, 50 \mathrm{~Hz}\), AC is applied to a resistor. The instantaneous value of voltage is
(a) Here, \(V_{\mathrm{rms}}=220 \mathrm{~V}, f=50 \mathrm{~Hz}\)
Peak value of voltage, \(V_0=\sqrt{2} V_{\text {rms }}=220 \sqrt{2} \mathrm{~V}\)
The instantaneous value of voltage is
\(
\begin{aligned}
V & =V_0 \sin 2 \pi f t=220 \sqrt{2} \sin 2 \pi \times 50 t \\
& =220 \sqrt{2} \sin 100 \pi t
\end{aligned}
\)
Frequency of AC mains in India is
(c) The standard frequency of AC mains in India is 50 Hz .
The instantaneous current in an AC circuit is \(I=\sqrt{2} \sin (50 t+\pi / 4)\). The rms value of current is
\(
\text { (d) } I_{\mathrm{rms}}=\frac{I_0}{\sqrt{2}}=\frac{\sqrt{2}}{\sqrt{2}}=1 \mathrm{~A}
\)
The peak value of an alternating current is 5 A and its frequency is 60 Hz . Find its rms value and time taken to reach the peak value of current starting from zero.
(a) \(\)I_{\mathrm{rms}}=\frac{I_0}{\sqrt{2}}=\frac{5}{\sqrt{2}}=3.536 \mathrm{~A}latex]
Time taken, \(\)T=\frac{1}{4 \times 60}=4.167 \mathrm{~ms}latex]
If an alternating voltage is represented as \(E=141 \sin (628 t)\), then the rms value of the voltage and the frequency are respectively
\(
\begin{aligned}
& \text { (c) } E=141 \sin (628 t) \\
& \quad E_{\mathrm{rms}}=\frac{E_0}{\sqrt{2}}=\frac{141}{1.41}=100 \mathrm{~V} \text { and } 2 \pi f=628 \\
& \Rightarrow \quad f=100 \mathrm{~Hz}
\end{aligned}
\)
An alternating current in a circuit is given by \(I=20 \sin (100 \pi t+0.05 \pi) \mathrm{A}\). The rms value and the frequency of current respectively, are
\(
\begin{aligned}
&\text { (c) } I=20 \sin (100 \pi t+0.05 \pi)\\
&\begin{aligned}
\therefore I_{\mathrm{rms}} & =\frac{20}{\sqrt{2}}=10 \sqrt{2} \mathrm{~A} \\
\omega & =100 \pi \Rightarrow f=50 \mathrm{~Hz}
\end{aligned}
\end{aligned}
\)
Ohm’s law expressed as \(E=I R\)
(c) Ohm’s law, expressed as \(E=I R\), applies to AC circuits when impedance \((Z)\) is substituted for resistance \((R)\). While it’s true that Ohm’s law ( \(\mathrm{V}=\mathrm{IR}\) ) is foundational in DC circuits, it can be adapted for AC circuits by using impedance, which accounts for both resistance and reactance (opposition to current flow due to capacitance or inductance.
An alternating current of rms value 10 A is passed through a \(12 \Omega\) resistor. The maximum potential difference across the resistor is
\(
\begin{aligned}
&\text { (c) Maximum potential difference, } V_0=I_0 R\\
&=\left(I_{\mathrm{rms}} \sqrt{2}\right) R=(10 \sqrt{2})(12)=169.70 \mathrm{~V}
\end{aligned}
\)
The reactance of a \(25 \mu \mathrm{~F}\) capacitor at the AC frequency of 4000 Hz is
\(
\begin{aligned}
&\text { (a) Capacitive reactance, }\\
&X_C=\frac{1}{2 \pi f C}=\frac{1}{2 \pi \times 4000 \times 25 \times 10^{-6}}=\frac{5}{\pi} \Omega
\end{aligned}
\)
The capacity of a pure capacitor is 1 F. In DC circuits, its effective resistance will be
\(
\begin{aligned}
&\text { (b) In DC circuits, } f=0\\
&X_C=\frac{1}{2 \pi(0) C}=\frac{1}{0}=\infty
\end{aligned}
\)
In an AC circuit, an alternating voltage \(e=200 \sqrt{2} \sin 100 t \mathrm{~V}\) is connected to a capacitor of capacity \(1 \mu \mathrm{~F}\). The rms value of the current in the circuit is
(c) Given, \(e=200 \sqrt{2} \sin 100 t\)
and \(\quad C=1 \mu \mathrm{~F}\)
Here, \(\quad E_{\mathrm{rms}}=\frac{200 \sqrt{2}}{\sqrt{2}}=200 \mathrm{~V}\)
\(
\begin{aligned}
X_C & =\frac{1}{\omega C}=\frac{1}{1 \times 10^{-6} \times 100}=10^4 \Omega \\
i_{\mathrm{rms}} & =\frac{E_{\mathrm{rms}}}{X_C} \Rightarrow i_{\mathrm{rms}}=\frac{200}{10^4}=2 \times 10^{-2} \mathrm{~A}=20 \mathrm{~mA}
\end{aligned}
\)
The reactance of a coil when used in the domestic AC power supply ( \(220 \mathrm{~V}, 50\) cycles per second) is \(50 \Omega\). The inductance of the coil is nearly
(d)
\(
\begin{aligned}
X_L & =(2 \pi f L) \\
L & =\frac{X_L}{2 \pi f}=\frac{50}{(2 \pi)(50)}=0.16 \mathrm{H}
\end{aligned}
\)
An inductive coil has a resistance of \(100 \Omega\). When an AC signal of frequency 1000 Hz is applied to the coil, the voltage leads the current by \(45^{\circ}\). The inductance of the coil is
\(
\begin{aligned}
& \text { (b) As, } \tan \phi=\frac{X_L}{R} \\
& \therefore \tan 45^{\circ}=\frac{L \omega}{R}
\end{aligned}
\)
\(
L=\frac{R}{\omega}=\frac{100}{2 \pi \times 1000} \quad\left[\because \tan 45^{\circ}=1\right]
\)
\(
L=\frac{1}{20 \pi}
\)
Two inductors \(L_1\) and \(L_2\) are connected in parallel and a time varying current flows as shown in figure. Then the ratio of currents \(I_1 / I_2\) at any time \(t\) is
\(
\begin{aligned}
& \text { (b) As, } \quad I \propto \frac{1}{\omega L} \\
& \therefore \frac{I_1}{I_2}=\frac{\omega L_2}{\omega L_1}=\frac{L_2}{L_1}
\end{aligned}
\)
An inductance and a resistance are connected in series with an AC potential. In this circuit,
(b) Here inductance and resistance are connected in series. We know that in case of resistance, both current and potential difference are in the same phase. In inductor, voltage leads current by \(\pi / 2\).
In \(L-R\) circuit, resistance is \(8 \Omega\) and inductive reactance is \(6 \Omega\), then impedance is
\(
\text { (d) } Z=\sqrt{R^2+X_L^2}=\sqrt{8^2+6^2}=10 \Omega
\)
In an AC circuit, the current lags behind the voltage by \(\frac{\pi}{3}\). The components in the circuit may be
(a) \(V_L=i X_L\) and \(V_R=i R\), where \(X_L\) is the inductive reactance. Now, \(V_R\) is in phase with the current i, while \(V_L\) leads i by \(90^{\circ}\).
The phasor diagram above shows that, in L-R circuit, the voltage leads the current by a phase angle \(\phi\) is given by:
\(
\begin{aligned}
& \tan \phi=\frac{V_L}{V_R}=\frac{i X_L}{i R}=\frac{\omega L}{R} \\
& \phi=\tan ^{-1} \frac{\omega L}{R}
\end{aligned}
\)
So, if in an AC circuit, the current lags behind the voltage by \(\pi / 3\). The components of the circuit are L and R.
In an AC circuit, a resistance \(R\) is connected in series with an inductance \(L\). If phase angle between voltage and current be \(45^{\circ}\), the value of inductive reactance will be
\(
\text { (c) } \tan 45^{\circ}=\frac{X_L}{R} \Rightarrow X_L=R
\)
In a circuit containing \(R\) and \(L\), as the frequency of the impressed AC increases, the impedance of the circuit
(b) Impedance, \(Z=\sqrt{R^2+(2 \pi f L)^2}\)
As \(f\) increases, \(Z\) will increase.
An AC voltage is applied to a resistance \(R\) and an inductor \(L\) in series. If \(R\) and the inductive reactance are both equal to \(3 \Omega\), the phase difference (in rad) between the applied voltage and the current in the circuit is
(a)
\(
\tan \phi=\frac{X_L}{R}=\frac{L \omega}{R}
\)
\(
\begin{aligned}
\tan \phi & =\frac{3}{3} \quad \Rightarrow \quad \tan \phi=1 \\
\phi & =\tan ^{-1}(1) \quad \Rightarrow \quad \phi=45^{\circ} \\
\phi & =\frac{\pi}{4} \mathrm{rad}
\end{aligned}
\)
In an \(L-R\) circuit, the value of \(L\) is \(\left(\frac{0.4}{\pi}\right) \mathrm{H}\) and the value of \(R\) is \(30 \Omega\). If in the circuit, an alternating emf of 200 V at 50 cycle s \(^{-1}\) is connected, the impedance of the circuit and current will be
\(
\text { (d) Given, } X_L=\omega L=2 \pi f L=2 \pi \times 50 \times \frac{0.4}{\pi}=40 \Omega
\)
\(
\begin{array}{ll}
\text { and } & R=30 \Omega \\
\therefore & Z=\sqrt{R^2+X_L^2}=\sqrt{(30)^2+(40)^2}=50 \Omega \\
\text { and } & I_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{Z}=\frac{200}{50}=4 \mathrm{~A}
\end{array}
\)
The instantaneous values of current and voltage in an AC circuit are given by
\(
\begin{aligned}
I & =6 \sin \left(100 \pi t+\frac{\pi}{4}\right) \\
V & =5 \sin \left(100 \pi t-\frac{\pi}{4}\right), \text { then }
\end{aligned}
\)
(c) The phase difference between instantaneous value of \(I\) and \(V\) is
\(
\frac{\pi}{4}-\left(-\frac{\pi}{4}\right)=\frac{\pi}{2}
\)
Hence, current leads the voltage by \(90^{\circ}\).
In an \(L-C-R\) series circuit the AC voltage across \(R, L\) and \(C\) come out as \(10 \mathrm{~V}, 10 \mathrm{~V}\) and 20 V, respectively. The voltage across the entire combination will be
\(
\text { (d) } V=\sqrt{V_R^2+\left(V_L-V_C\right)^2}=10 \sqrt{2} \mathrm{~V}
\)
With increase in frequency of an AC supply, the impedance of an \(L-C-R\) series circuit
(d) Impedance (Z) in an L-C-R series circuit is defined as the total opposition to the flow of alternating current. It is given by the formula:
\(
Z=\sqrt{R^2+\left(X_L-X_C\right)^2}
\)
where \(R\) is the resistance, \(X_L\) is the inductive reactance, and \(X_C\) is the capacitive reactance.
The inductive reactance \(X_L\) is given by:
\(
X_L=2 \pi f L
\)
The capacitive reactance \(X_C\) is given by:
\(
X_C=\frac{1}{2 \pi f C}
\)
where \(f\) is the frequency, \(L\) is the inductance, and \(C\) is the capacitance.
At \(f=0\) :
\(X_L=0\)
\(X_C\) approaches infinity.
Thus, impedance \(Z\) is infinite.
As the frequency \(f\) increases:
\(X_L\) increases (since it is directly proportional to \(f\) ).
\(X_C\) decreases (since it is inversely proportional to \(f\) ).
Initially, as frequency increases, \(Z\) decreases because \(X_C\) decreases faster than \(X_L\) increases.
There exists a frequency where \(X_L=X_C\). At this point:
\(
Z=R
\)
This is the minimum impedance of the circuit.
Beyond the resonance frequency:
\(X_L\) continues to increase.
\(X_C\) continues to decrease.
Eventually, \(Z\) starts to increase again because the increase in \(X_L\) outweighs the decrease in \(X_C\).
As \(f\) approaches infinity:
\(X_L\) approaches infinity.
\(X_C\) approaches zero.
Thus, impedance \(Z\) becomes infinite again.
The impedance of the L-C-R circuit first decreases with increasing frequency, reaches a minimum value at resonance, and then increases again as frequency continues to rise.
A sinusoidal voltage of peak value 300 V and an angular frequency \(\omega=400 \mathrm{rads}^{-1}\) is applied to series \(L-C-R\) circuit, in which \(R=3 \Omega, L=20 \mathrm{mH}\) and \(C=625 \mu \mathrm{~F}\). The peak value of current in the circuit is
(b) The impedance of the circuit is
Here,
\(
\begin{aligned}
Z & =\sqrt{R^2+\left(X_L-X_C\right)^2} \\
X_L & =\omega L=400 \times 20 \times 10^{-3}=8 \mathrm{H}
\end{aligned}
\)
\(
\begin{array}{ll}
\text { and } & X_C=\frac{1}{\omega C}=\frac{1}{400 \times 625 \times 10^{-6}}=4 \mathrm{~F} \\
\therefore & Z=\sqrt{(3)^2+(8-4)^2}=5 \Omega \\
\therefore & I=\frac{E}{Z}=\frac{300}{5}=60 \mathrm{~A}
\end{array}
\)
The value of current at resonance in a series \(L-C-R\) circuit is affected by the value of
(a) In a series \(\boldsymbol{L}-\boldsymbol{C}-\boldsymbol{R}\) circuit at resonance, the impedance \((\mathrm{Z})\) is minimized, and this minimum impedance is equal to the resistance (R) of the circuit. The current (I) in the circuit at resonance is given by the formula \(\boldsymbol{I}=\frac{\boldsymbol{V}}{\boldsymbol{R}}\), where V is the applied voltage.
An \(L-C-R\) series circuit is connected to a source of alternating current. At resonance the applied voltage and current flowing through the circuit will have a phase difference of
(a) In a series LCR circuit the phase angle is given by \(\tan \phi=\frac{\omega L-\frac{1}{N}}{R}\)
At resonance \(\omega L=\frac{1}{\omega C}\), then \(\tan \phi=0\), i.e., \(\phi=0\)
A series \(L-C-R\) circuit is operated at resonance. Then
(b) The impedance \((Z)\) of a series \(L-C-R\) circuit is given by the formula:
\(
Z=\sqrt{R^2+\left(X_L-X_C\right)^2}
\)
Since at resonance \(X_L=X_C\), the equation simplifies to:
\(
Z=\sqrt{R^2+0^2}=R
\)
This indicates that the impedance is at its minimum value, which is equal to the resistance (R).
An \(L-C-R\) series circuit is under resonance. If \(I_m\) is current amplitude, \(V_m\) is voltage amplitude, \(R\) is the resistance, \(Z\) is the impedance, \(X_L\) is the inductive reactance and \(X_C\) is the capacitive reactance, then
(d) Impedance of the circuit, \(Z=\sqrt{R^2+\left(X_L-X_C\right)^2}\)
At resonance, \(X_L=X_C\)
\(
\begin{aligned}
& \therefore Z=R \\
& \therefore I_m=\frac{V_m}{Z} \\
& =\frac{V_m}{R}
\end{aligned}
\)
In an \(L-C-R\) series AC circuit, at resonance
(b) At resonance (series resonant circuit),
\(
X_L=X_C \Rightarrow Z_{\min }=R
\)
Impedance is minimum and current is maximum.
An \(L-C-R\) series circuit connected to a source \(E\), is at resonance. Then,
(b) Equation of voltage, \(E=\sqrt{V_R^2+\left(V_L-V_C\right)^2}\)
At resonance (series circuit),
\(
V_L=V_C \Rightarrow E=V_R
\)
i.e. Whole applied voltage appeared across the resistance.
An electric heater rated 220 V and 550 W is connected to AC mains. The current drawn by it is
(b)
\(
\begin{aligned}
P & =V I \\
I & =\frac{P}{V}=\frac{550}{220}=2.5 \mathrm{~A}
\end{aligned}
\)
In an AC circuit, the potential difference \(V\) and current \(I\) are given respectively by
\(
\begin{aligned}
& V=100 \sin (100 t) \mathrm{V} \\
& I=100 \sin \left(100 t+\frac{\pi}{3}\right) \mathrm{mA}
\end{aligned}
\)
The power dissipated in the circuit will be
(c) Voltage amplitude, \(V_0=100 \mathrm{~V}\), current amplitude \(I_0=100 \mathrm{~mA}=100 \times 10^{-3} \mathrm{~A}\) and phase difference between \(I\) and \(V\) is \(\phi=\frac{\pi}{3}=60^{\circ}\).
Now power dissipated is given by
\(
\begin{aligned}
P & =\frac{V_0 I_0}{2} \cos \phi \\
& =\frac{100 \times 100 \times 10^{-3}}{2} \times \cos 60^{\circ}=2.5 \mathrm{~W}
\end{aligned}
\)
In an AC circuit, the power factor
(b) Power factor of an AC circuit is given by
\(
\cos \phi=\frac{R}{Z}
\)
For circuit containing an ideal resistance only, \(Z=R\)
\(
\therefore \quad \cos \phi=1
\)
The average power dissipated in a pure inductor of inductance \(L\), when an AC current is passing through it, is
(d) In case of a pure inductance, \(R=0\)
\(\therefore \quad\) Power factor \(=\cos \phi=\frac{R}{Z}=\frac{0}{X}=0\)
So, \(\quad P_{\mathrm{av}}=E_{\mathrm{rms}} \cdot i_{\mathrm{rms}} \times 0=0\)
The average power dissipated in a pure capacitance \(C\) in an AC circuit is
(b) The average power in an AC circuit is given by \(P_{\text {avg }}=V_{\text {rms }} I_{\text {rms }} \cos (\phi)\).
For a pure capacitor, the voltage lags the current by a phase angle of \(\phi=\frac{\pi}{2}\) radians.
The average power dissipated is \(P_{\text {avg }}=V_{\text {rms }} I_{\text {rms }} \cos (\phi)\).
Substitute the power factor: \(P_{\text {avg }}=V_{\text {rms }} I_{\text {rms }} \times 0\).
Therefore, \(P_{\text {avg }}=0\).
The impedance of a circuit consists of \(3 \Omega\) resistance and \(4 \Omega\) reactance. The power factor of the circuit is
\(
\begin{array}{r}
\text { (b) } Z=\sqrt{(3)^2+(4)^2}=5 \Omega \\
\cos \phi=\frac{R}{Z}=\frac{3}{5}=0.6
\end{array}
\)
Power dissipated in an \(L-C-R[latex] series circuit connected to an AC source of emf [latex]\varepsilon\) is
(a)
\(
\begin{aligned}
&\text { The impedance } Z \text { for an } L-C-R \text { series circuit is: }\\
&Z=\sqrt{R^2+\left(L \omega-\frac{1}{C \omega}\right)^2}
\end{aligned}
\)
The RMS current \(I_{r m s}\) is given by the RMS emf divided by the impedance:
\(
I_{r m s}=\frac{\varepsilon_{r m s}}{Z}
\)
Assuming \(\varepsilon\) represents the RMS emf, then:
\(
I_{r m s}=\frac{\varepsilon}{\sqrt{R^2+\left(L \omega-\frac{1}{C \omega}\right)^2}}
\)
The power dissipated \(P\) in an AC circuit is given by:
\(
P=I_{r m s}^2 R
\)
Substitute the expression for \(I_{\text {rms }}\) :
\(
P=\left(\frac{\varepsilon}{\sqrt{R^2+\left(L \omega-\frac{1}{C \omega}\right)^2}}\right)^2 R
\)
Simplify the expression:
\(
P=\frac{\varepsilon^2 R}{R^2+\left(L \omega-\frac{1}{C \omega}\right)^2}
\)
The energy stored in an inductor of self-inductance \(L\) henry carrying a current of \(I\) is
\(
\text { (b) The energy stored in an inductor, } U=\frac{1}{2} L I^2
\)
In an inductor of inductance \(L=100 \mathrm{mH}\), a current of \(I=10 \mathrm{~A}\) is flowing. The energy stored in the inductor is
\(
\text { (a) } U=\frac{1}{2} L I^2=\frac{1}{2}\left(100 \times 10^{-3}\right)(10)^2=5 \mathrm{~J}
\)
Choke coil is a device having
(d) A choke coil is a device with high inductance and low resistance.
Explanation:
Choke coils are designed to offer high impedance to alternating current (AC) while allowing direct current (DC) to pass with minimal resistance. This is achieved by using a coil with a high inductance and a low resistance.
High inductance:
This ensures that the choke coil offers a high impedance to AC, effectively blocking or reducing the flow of AC while allowing DC to pass.
Low resistance:
This minimizes power loss in the form of heat, which is crucial for efficient circuit operation.
What is increased in step-down transformer?
(b) In step-down transformer, \(V_P>V_S\)
\(
\begin{array}{ll}
\text { but, } & \frac{V_P}{V_S}=\frac{I_s}{I_P} \\
\therefore & I_S>I_P
\end{array}
\)
So, current increases in a step-down transformer.
A transformer works on the principle of
(c) A transformer works on the principle of mutual induction.
Explanation: Mutual induction refers to the process where a changing current in one coil creates a magnetic field that induces a voltage in a nearby coil, which is the fundamental mechanism behind how transformers operate.
Quantity that remains unchanged in a transformer is
(c) The quantity that remains unchanged in a transformer is frequency.
Explanation: Transformers work by changing the voltage and current levels, but they do not alter the frequency of the AC power supplied to them.
The ratio of secondary to the primary turns in a transformer is \(3: 2\). If the power output be \(P\), then the input power neglecting all losses must be equal to
(c) If we neglect all losses, then input power = output power.
The transformation ratio in the step-up transformer is
(b) The transformation ratio, \(K\), is given by \(K=\frac{V_s}{V_p}=\frac{N_s}{N_p}\).
Here, \(\boldsymbol{V}_s\) is the secondary voltage, \(\boldsymbol{V}_{\boldsymbol{p}}\) is the primary voltage.
Also, \(N_s\) is the number of turns in the secondary coil, and \(N_p\) is the number of turns in the primary coil.
In a step-up transformer, the secondary voltage is greater than the primary voltage, so \(V_s>V_p\).
This also means the number of turns in the secondary coil is greater than in the primary coil, so \(N_s>N_p\).
Since \(V_s>V_p\) (or \(N_s>N_p\) ), the ratio \(\frac{V_s}{V_p}\) (or \(\frac{N_s}{N_p}\) ) must be greater than 1 . Therefore, \(K>1\).
The transformation ratio in a step-up transformer is greater than one.
In a transformer, the number of turns in primary and secondary are 500 and 2000, respectively. If current in primary is 48 A, the current in the secondary is
\(
\text { (a) As, } \frac{N_1}{N_2}=\frac{I_2}{I_1} \Rightarrow I_2=\left(\frac{N_1}{N_2}\right) I_1=\left(\frac{500}{2000}\right) \times 48=12 \mathrm{~A}
\)
The core used in a transformer and other electromagnetic devices is laminated so that
(b) The core used in a transformer and other electromagnetic devices is laminated to energy loss due to eddy currents may be minimised.
Explanation: Laminating the core creates thin layers of metal separated by insulation, which significantly reduces the flow of eddy currents within the core, thereby minimizing energy loss as heat.
Why other options are incorrect:
(a) ratio of voltage in the primary and secondary may be increased:
The ratio of voltages in a transformer is determined by the number of turns in the primary and secondary coils, not the lamination of the core.
(c) the weight of the transformer may be reduced:
While laminating can slightly increase the size of the core due to the added insulation layers, it doesn’t significantly reduce the weight.
(d) residual magnetism in the core may be reduced:
While laminating can help with reducing eddy currents, it’s not the primary method for reducing residual magnetism. Special materials and proper demagnetization techniques are used for that purpose.
Which of the following is constructed on the principle of electromagnetic induction?
(c) The device constructed on the principle of electromagnetic induction is a generator.
Explanation: Electromagnetic induction refers to the generation of an electric current in a conductor when it is exposed to a changing magnetic field. A generator uses this principle by rotating a coil of wire within a magnetic field, causing a current to flow through the coil.
Why other options are incorrect:
Galvanometer:
A galvanometer detects and measures electric current by using the magnetic field produced by the current, but it does not rely on the principle of electromagnetic induction where a changing magnetic field induces a current.
Electric motor:
An electric motor works by converting electrical energy into mechanical energy, using the magnetic force exerted on a current-carrying conductor within a magnetic field. While it involves magnetic fields, it does not use the principle of electromagnetic induction in the same way as a generator.
Voltmeter:
A voltmeter measures the potential difference (voltage) in a circuit. It does not use the principle of electromagnetic induction, but rather relies on Ohm’s law.
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