Question

An alternating emf \(E = 440 \sin(100\pi t)\) is applied to a circuit containing an inductance of \(\frac{\sqrt{2}}{\pi}\) H. If an a.c. ammeter is connected in the circuit, its reading will be:

• A. 4.4 A
• B. 1.55 A
• C. 2.2 A
• D. 3.11 A

Answer 1

The Correct Option is C

To determine the reading on an AC ammeter connected to a circuit, we need to find the current \( I \) flowing through the circuit when the alternating EMF \( E \) is applied across an inductor.

The given EMF is:

\(E = 440 \sin(100\pi t)\) 

The formula for current in an inductor when alternating EMF is applied is given by:

\(I = \frac{E_0}{Z}\)

where \( E_0 \) is the peak EMF and \( Z \) is the impedance of the inductor.

Since there is only an inductor in the circuit, the impedance \(Z = \omega L\), where:

• \(\omega\) is the angular frequency.
• \(L\) is the inductance.

From the given equation of EMF:

\(\omega = 100\pi\)

Given, the inductance \(L = \frac{\sqrt{2}}{\pi}\) H.

Therefore, the impedance:

\(Z = \omega L = 100\pi \times \frac{\sqrt{2}}{\pi} = 100\sqrt{2}\)

The peak EMF \( E_0 \) is given as 440 V.

Using the formula for current:

\(I = \frac{440}{100\sqrt{2}}\)

Calculate the current:

\(I = \frac{440}{100\sqrt{2}} = \frac{440}{141.42} \approx 3.11 \text{ A}\)

Since the ammeter measures the RMS value of the current, and for a sine wave:

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

Therefore, RMS current:

\(I_{\text{rms}} = \frac{3.11}{\sqrt{2}} = 2.2 \text{ A}\)

Thus, the reading on the AC ammeter will be 2.2 A, which matches the correct given option.

Answer 2

Current \(I = \frac{V}{\omega L}\)
\(I = \frac{440}{100\pi \times \frac{\sqrt{2}}{\pi}}\)
\(I = \frac{44}{10\sqrt{2}}\)
\(⇒\) \(I_{\text{rms}} = \frac{I}{\sqrt{2}}\)
\(=\frac{44}{20}\)
\(=2.2 A\)
So, the correct option is (C): 2.2 A