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

Answer: (C)

2.2 A

Answer 2

$i(t) = \frac{V}{R} \left(1 - e^{-\frac{Rt}{L}}\right) \quad \dots (1)$
$\frac{L}{R} = \frac{1}{100s}$

$⇒$ $\frac{L}{R} = 10 \ \text{ms} \quad \dots (2)$
$\frac{V}{2R} = \frac{V}{R} \left(1 - e^{-\frac{Rt}{L}}\right)$

$⇒$ $e^{-\frac{Rt}{L}} = \frac{1}{2}$

$⇒$ $t = \frac{L}{R} \ln 2 = 6.93 \ \text{ms}$

$U = \frac{1}{2}Li^2$
$= \frac{1}{2}[1 - e^{-\frac{15}{10}}]^2 \left(\frac{6}{100}\right)^2$
=$$\frac{1}{2}[1 - 0.25]^2 \times 36 \times 10^{-4}
$$$=$ 1 mJ
So, the correct option is (C): t = 7 ms; U = 1 mJ