Question

In the given circuit, $R$ is a pure resistor, $L$ is a pure inductor, S is a 100V, 50Hz AC source and A is an AC ammeter. With either $K_1$ or $K_2$ alone closed, the ammeter reading is $I$. If the source is changed to 100 V, 100 Hz, the ammeter reading with $K_1$ alone closed and with $K_2$ alone closed will be respectively

• A. $I,\frac{I}{2}$
• B. $I,2I$
• C. $2I,I$
• D. $2I,\frac{I}{2}$

Answer 1

The correct option is A $I,\frac{I}{2}$

$I=\frac{100}{R}=\frac{100}{2\pi \left(50\right)\left(L\right)}$

now at $100Hz$

new I when $K_1$ is close only

$I_1=\frac{100}{R}=I$

and new I when only $K_2$ is closed

$I_2=\frac{100}{2\pi \times \left(100\right)\times L}=\frac{I}{2}$