Question

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

• A. 2 A
• B. 4 A
• C. 3 A
• D. 5 A

Hint:

In RL circuits, the current can be determined using the equation \( \epsilon = L \frac{dI}{dt} + IR \). Make sure to substitute the correct values and solve for the unknown current.

Answer 1

The Correct Option is C

Using the formula:
\[ \text{Battery Voltage} - \text{Inductor Voltage} - \text{Resistor Voltage} = 0 \] \[ 12 - L\frac{dI}{dt} - IR = 0 \]

We are given:

Battery Voltage = 12V

Inductance (L) = 3H

\(\frac{dI}{dt} = -8\) A/s (current is decreasing)

Resistance (R) = 12Ω

Plugging these values in:

\[ 12 - 3(-8) - 12I = 0 \] \[ 12 + 24 - 12I = 0 \] \[ 36 - 12I = 0 \] \[ 12I = 36 \] \[ I = \frac{36}{12} = 3 \] Thus, the current (I) is 3 A.  Answer: 3 A