M.C.Q (1 Marks)

Question 11 Mark

An AC source rated 100V (rms) supplies a current of 10A (rms) to a circuit. The average power delivered by the source:

Must be 1000W.
May be 1000W.
May be greater than 1000W.
May be less than 1000W.

Answer

May be 1000W.

May be less than 1000W.

Explanation:
Average power $\text{P}_\text{av}=\text{V}_\text{rms}\text{I}_\text{rms}\cos\phi$
$=100\times10\cos\phi$
$\text{P}_\text{av}=1000\cos\phi$
$\therefore\ \cos\phi$ lies "0 to 1".
$\Rightarrow\ 0\leq\text{P}_\text{av}\leq1000.$

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Question 21 Mark

An AC source is rated 220V, 50Hz. The average voltage is calculated in a time interval of 0.01s, It:

Must be zero.
May be zero.
Is never zero.
Is $\Big(\frac{200}{\sqrt{2}}\Big)\text{V}.$

Answer

May be zero.

Explanation:

$\text{V}=\text{V}_0\sin\omega\text{t}$

$\omega=2\pi\text{f}=2\times3.14\times50$
$\omega=314$
$\text{V}_\text{avg}=\frac{\int\limits_0^{0.01}\text{V}\text{dt}}{\int\limits_0^{0.01}\text{dt}}$
$=\text{V}_0\Big(\frac{1\cos\omega\text{t}}{\omega}\Big)_0^{0.01}$
$=\frac{\text{V}_0}{\omega\times0.01}\big(1-\cos\omega(0.1)\big)$
$=\frac{\text{V}_0}{314\times0.01}\big(1-\cos(314\times0.01)\big)$
$=\frac{\text{V}_0}{3.14}\big(1-\cos(314)\big)$
$=\frac{\text{V}_0}{3.14}\big(1-\cos\pi\big)$
$=\frac{2\text{V}_0}{\pi}=140.127\text{volt}$

if $\text{V}=\text{V}_0\cos\omega\text{t}$
$\text{V}_\text{avg}=\frac{\int\text{V d}\rho}{\int\text{dt}}=0$

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Question 31 Mark

In an AC series circuit, the instantaneous current is zero when the instantaneous voltage is maximum. Connected to the source may be a:

Pure inductor.
Pure capacitor.
Pure resistor.
Combination of an inductor and a capacitor.

Answer

Pure inductor.
Pure capacitor.

Combination of an inductor and a capacitor.

Explanation:
Instantaneous current is zero when the intantaneous voltage is maximum.
Mean resistance = 0.

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Question 41 Mark

The magnetic field energy in an inductor changes from maximum value to minimum value in 5.0ms when connected to an AC source. The frequency of the source:

20Hz.
50Hz.
200Hz.
500Hz.

Answer

50Hz.

Explanation:
Frequency of the source is remain constant = 50Hz.

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MCQ 51 Mark

An $AC$ source producing emf $\epsilon=\epsilon_{0}\Big[\cos\big(100\pi\text{s}^{-1}\big)\text{t}+\cos\big(500\pi\text{s}^{-1}\big)\text{t}\Big]$ is connected in series with a capacitor and a resistor. The steady$-$state current in the circuit is found to be $\text{i}=\text{i}_1\cos\Big[\big(100\pi\text{s}^{-1}\big)\text{t}+\phi_1\Big]+\text{i}_2\cos\Big[\big(500\pi\text{s}^{-1}\big)\text{t}+\phi_2\Big].$

A
$i_1 > i_2$
B
$i_1 = i_2$
✓
$i_1 < i_2$
D
The information is insufficient to find the relation between $i_1$ and $i_2$

Answer

Correct option: C.

$i_1 < i_2$

$\text{Q}=\text{C}\epsilon=\epsilon_{0}\text{C}\Big[\cos\big(100\pi\text{s}^{-1}\big)\text{t}+\cos\big(500\pi\text{s}^{-1}\big)\text{t}\Big]$
$\text{i}=\frac{\text{dQ}}{\text{dt}}$
$\text{Q}=\text{C}\epsilon=\epsilon_{0}\text{C}\Big[\cos\big(100\pi\text{s}^{-1}\big)\text{t}+\cos\big(500\pi\text{s}^{-1}\big)\text{t}\Big]$
$\epsilon_0\text{C}\times100\pi\Big[\sin\big(100\pi\text{s}^{-1}\big)\text{t}\Big]\\+\epsilon_0\text{C}\times500\pi\Big[\sin\big(500\pi\text{s}^{-1}\big)\text{t}\Big]$
$=100\text{C}\pi\epsilon_0\cos\Big[\big(100\pi\text{s}^{-1}\big)\text{t}+\phi_1\Big]\\+500\text{C}\pi\epsilon_0\cos\Big[\big(500\pi\text{s}^{-1}\big)\text{t}+\phi_2\Big]$
$\text{i}_1=100\pi\epsilon_0\text{C}$ and $\text{i}_2=500\pi\epsilon_0\text{C}$
$\text{i}_2>\text{i}_1$

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Question 61 Mark

A series AC circuit has a resistance of $4\Omega$ and a reactance of $3\Omega.$ The impedance of the circuit is:

$5\Omega$
$7\Omega$
$\frac{12}{7}\Omega$
$\frac{7}{12}\Omega$

Answer

$5\Omega$

Explanation:
$\text{Z}=\sqrt{\text{R}^2+\text{X}^2}$
$\text{R}=4\Omega,\text{X}=3\Omega$
$=\text{Z}=\sqrt{4^2+3^2}=5\Omega$

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Question 71 Mark

The peak voltage in a 220V AC source is:

220V.
About 160V.
About 310V.
440V.

Answer

About 310V.

Explanation:
$\text{V}_\text{rms}=220\text{V}$
$\text{V}_\text{p}=\sqrt{2}\times\text{V}_\text{rms}$
$=220\times1.414=311\text{volt}$

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Question 81 Mark

Transformers are used:

In DC circuits only.
In AC circuits only.
In both DC and AC circuits.
Neither in DC nor in AC circuits.

Answer

In AC circuits only.

Explanation:
Transformers are used in AC circuits only.

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Question 91 Mark

The reactance of a circuit is zero. It is possible that the circuit contains:

An inductor and a capacitor.
An inductor but no capacitor.
A capacitor but no inductor.
Neither an inductor nor a capacitor.

Answer

An inductor and a capacitor.

Explanation:
$\text{X}=0$ (Given)
$\text{X}=\text{X}_\text{L}+\text{X}_\text{C}$
$=\omega\text{L}-\frac{1}{\omega\text{C}}=0$
It is possible that the circuit contains an inductor and a capacitor.

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Question 101 Mark

An inductor coil of some resistance is connected to an AC source. Which of the following quantities have zero average value over a cycle?

Current.
Induced emf in the inductor.
Joule heat.
Magnetic energy stored in the inductor.

Answer

Current.
Induced emf in the inductor.

Explanation:
$\text{I}=\text{I}_0\sin\omega\text{t}$

Average value of current over a cycle = 0
$\text{V}=\text{V}_0\cos\omega\text{t}$
$=\text{V}_0\sin\bigg(\omega\text{t}+{\frac{\pi}{2}}\bigg)$

Average value of induced emf in inductor over a cycle = 0.

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MCQ 111 Mark

An inductor, a resistance 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.
B
Of the resistor increases.
C
Of the capacitor increases.
D
Of the circuit increases.

Answer

Correct option: A.

Of the inductor increases.

$\text{X}_\text{L}=\omega\text{L}$
$\text{X}_\text{C}=\frac{1}{\omega\text{C}}$
If frequency increases that causes $'X_L\ '$ raction of inductor increases and $'X_C\ '$ reactance of capacitor decreses.

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Question 121 Mark

A capacitor acts as an infinite resistance for:

DC.
AC.
DC as well as AC.
Neither AC nor DC.

Answer

Explanation:
$\text{X}_\text{C}\frac{1}{\omega\text{C}}=\frac{1}{0\times\text{C}}$ $\bigg\{\text{in}\stackrel{{\text{DC}}}{{\omega = 0 }}\bigg\}$
$=\infty$

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Question 131 Mark

An alternating current is given by $\text{i}=\text{i}_1\cos\omega\text{t}+\text{i}_2\sin\omega\text{t}.$ The rms current is given by:

$\frac{\text{i}_1+\text{i}_2}{\sqrt{2}}$
$\frac{|\text{i}_1+\text{i}_2|}{\sqrt{2}}$
$\sqrt{\frac{\text{i}_1^2+\text{i}_2^2}{2}}$
$\sqrt{\frac{\text{i}_1^2+\text{i}_2^2}{\sqrt{2}}}$

Answer

$\sqrt{\frac{\text{i}_1^2+\text{i}_2^2}{2}}$

Explanation:
$\text{i}=\text{i}_1\cos\omega\text{t}+\text{i}_2\sin\omega\text{t}$
$\text{I}_\text{rms}=\frac{\int\limits_0^\text{T}\text{I}^2\text{dt}}{\int\limits_0^\text{T}\text{dt}}$
if $\text{I}=\cos\omega\text{t}$
$\text{I}_\text{rms}^2=\frac{\text{I}_0^2}{2}$
$\text{i}=\text{i}_1\cos\omega\text{t}+\text{i}_2\sin\omega\text{t}$
Than $\text{i}_\text{rms}^2=\frac{\text{i}_1^2}{2}+\frac{\text{i}_2^2}{2}$
$\text{i}_\text{rms}=\sqrt{\frac{\text{i}_1^2+\text{i}_2^2}{2}}$

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Question 141 Mark

To convert mechanical energy into electrical energy, one can use:

DC dynamo.
AC dynamo.
Motor.
Transformer.

Answer

DC dynamo.
AC dynamo.

Explanation:
DC dynamo or AC dynamo use to convert mechnical energy into electrial energy.

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Question 151 Mark

A constant current of 2.8A exists in a resistor. The rms current is:

2.8A.
About 2A.
1.4A.
Undefined for a direct current.

Answer

2.8A.

Explanation:
A constant current exists in a resistor is rms current it is equal to 2.8Amp.

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Question 161 Mark

The AC voltage across a resistance can be measured using:

A potentiometer.
A hot-wire voltmeter.
A moving-coil galvanometer.
A moving-magnet galvanometer.

Answer

A hot-wire voltmeter.

Explanation:
The AC voltae across a resustance can be measured using a hot-wore volmeter.

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