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.

Instantaneous resistance, $$R = 12 \ \Omega$$

Magnitude of the rate of change of current, $$\left| \frac{di}{dt} \right| = 8 \text{ A/s}$$

The problem states that the sliding contact is "pulled outwards". Looking at the variable resistor in the circuit diagram, pulling the contact outwards includes more of the resistive material in the closed loop, which means the total resistance $$R$$ is increasing.

Because the resistance is increasing, the current $$i$$ in the circuit must be decreasing over time. Therefore, the rate of change of current is negative:

$$ \frac{di}{dt} = -8 \text{ A/s} $$

Now, apply Kirchhoff's Voltage Law (KVL) to the circuit loop at this specific instant:

$$ V - L\frac{di}{dt} - iR = 0 $$

$$ 12 - 3(-8) - i(12) = 0 $$

$$ 12 + 24 - 12i = 0 $$

$$ 36 - 12i = 0 $$

$$ 12i = 36 $$

$$ i = 3 \text{ A} $$