Question

In the given circuit the sliding contact is pulled outwards such that electric current in the circuit changes at the rate of 8 A/s. At an instant when R is $12\Omega$, the value of the current in the circuit will be ______ A.

Answer 1

-1

Answer 2

For a series R–L circuit, applying KVL we get:

$V = IR + L\,\frac{dI}{dt}$

Given $V = 12\,\text{V}$, $R = 12\,\Omega$, $L = 3\,\text{H}$, and $\frac{dI}{dt} = 8\,\text{A/s}$, substitute the values:

$12 = 12I + 3(8)$

$12 = 12I + 24$

$12I = 12 - 24 = -12$

$I = \frac{-12}{12} = -1\,\text{A}$

Explanation (minimal):

• KVL gives: $12 = 12I + 3(8)$.
• Solve to find $I = -1$ A (negative sign indicates the actual current direction is opposite to the assumed direction).