← Back to Alternating Current

Average Current and RMS Current

Practice Questions:

Question 1: moderate

Match the following and choose the correct option from given codes :

1. i-p, ii-q, iii-r

2. i-q, ii-r, iii-p

3. i-q, ii-r, iii-q

4. None of these

View Answer

💬 Discuss this Question

Question 2: difficult

The current in a circuit varies with time as I = 2 √t . Then the rms value of the current for the interval t = 2 to t = 4 sec is

1. √3 A

2. 2 √3 A

3. √3 /2A

4. (4 - 2√2) A

View Answer

💬 Discuss this Question

Question 3: moderate

The r.m.s. value of potential difference V shown in the figure is :-

1. \[\frac{V_{0}}{\sqrt{3}}\]

2. \[V_{0}\]

3. \[\frac{V_{0}}{\sqrt{2}}\]

4. \[\frac{V_{0}}{{2}}\sqrt{\frac{5}{2}}\]

View Answer

💬 Discuss this Question

Question 4: moderate

A square wave current switching rapidly between \(+0.5\) ampere and \(-0.5\) ampere is passed through an A.C ammeter. Then the reading shown by it, is

1. 0.25 ampere

2. 0.5 ampere

3. \(\frac{0.5}{\sqrt{2}}\) ampere

\(0.5 \times \sqrt{2}\) ampere

View Answer

An AC ammeter measures the RMS value of the current. For a symmetric square wave of amplitude \(I_0\), the RMS value is equal to its peak value, \(I_{rms} = I_0 = 0.5\text{ A}\).

💬 Discuss this Question

Question 5: easy

The peak value of the alternating current given by \(I = 4 \sin\omega t + 4 \sin(\omega t + 2\pi/3)\) is:

1. \(4\sqrt{2}\)

2. \(2\sqrt{2}\)

3. \(8\)

4. \(4\)

View Answer

The two currents can be vectorially added with phase difference \(\phi = 2\pi/3 = 120^\circ\). Resultant amplitude is \(I_0 = \sqrt{4^2 + 4^2 + 2 \times 4 \times 4 \cos 120^\circ} = 4\).

💬 Discuss this Question

Question 6: easy

The peak voltage of the AC source is equal to

1. The value of voltage applied to the circuit

2. The rms value of the source

3. \(\sqrt{2}\) times the rms value of the AC source

4. \(\frac{1}{\sqrt{2}}\) times the average value of the AC source

View Answer

For a sinusoidal AC voltage source, the relation between peak voltage \(V_0\) and root-mean-square voltage \(V_{\text{rms}}\) is \(V_0 = \sqrt{2}V_{\text{rms}}\).

💬 Discuss this Question

Question 7: easy

Assertion (A): The moving coil ammeters or voltmeters can’t be employed to measure alternating current or voltage respectively.

Reason (R): If an alternating current is passed through a moving coil ammeter or voltmeter, then the average value of torque experienced by the coil is zero.

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)

2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)

3. (A) is true but (R) is false

4. Both (A) and (R) are false

View Answer

Moving coil instruments measure the average value of current. For AC, the average value over a full cycle is zero, resulting in zero average torque. Hence, moving coil ammeters/voltmeters cannot measure AC. Both assertion and reason are true, and the reason correctly explains the assertion.

💬 Discuss this Question

Question 8: easy

Assertion (A): \(220V\), \(50 \text{ Hz}\) appliance implies that potential difference in bulb is always \(220V\).

Reason (R): Every appliance is specified with its peak tolerable voltage.

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)

2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)

3. (A) is true but (R) is false

4. Both (A) and (R) are false

View Answer

\(220V\) AC represents the RMS voltage, not the instantaneous or peak voltage. The instantaneous voltage varies sinusoidally, reaching a peak of \(V_{\text{peak}} = V_{\text{RMS}} \sqrt{2} = 220 \sqrt{2} \approx 311 \text{V}\). So, Assertion (A) is false. Appliances are usually specified by their RMS operating voltage, not peak tolerable voltage. So, Reason (R) is false. Both assertion and reason are false.

💬 Discuss this Question

Question 9: easy

Assertion (A): Transformer does not work on \( \text{dc} \).

Reason (R): \( \text{dc} \) neither changes in magnitude nor in direction.

1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)

2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)

3. (3) (A) is true but (R) is false

4. (4) Both (A) and (R) are false

View Answer

Transformers operate on the principle of mutual induction, requiring a changing magnetic flux. \( text{dc} \) current produces a constant magnetic field, thus no change in flux and no induced EMF. Hence, (A) is true. \( text{dc} \) current indeed has constant magnitude and direction, so (R) is also true and correctly explains (A).

💬 Discuss this Question

Question 10: easy

Assertion (A): The divisions are equally marked on the scale of \( \text{ac} \) ammeter.

Reason (R): Heat produced is directly proportional to the current.

1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)

2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)

3. (3) (A) is true but (R) is false

4. (4) Both (A) and (R) are false

View Answer

\( \text{ac} \) ammeters (hot wire type) measure current based on the heating effect, where heat produced is \( H \propto I^2 \). This quadratic relationship results in a non-linear scale, not equally marked. Thus (A) is false. Heat produced is proportional to the square of the current, not directly, so (R) is also false.

💬 Discuss this Question