Question

In the circuit shown charge q varies with time t as \(q = t^2-5\), where q is in coulomb and t is in second. At time \(t = 3\) second, voltage \(V_{AB}\) in volt will be

• A. \(8\)
• B. \(12\)
• C. \(14\)
• D. \(18\)

Hint:

The current is dq/dt; add the drops across C, L and R along the current direction.

Answer 1

The Correct Option is C

Step 1: Use Kirchhoff style sum:
Going from A to B along the current: $V_A - V_B = \frac qC + L\frac{dI}{dt} + IR$, because each element drops potential in the direction of the current.

Step 2: Evaluate each at t = 3 s:
$q = 4$ C, so $\frac qC = 1$ V. $I = 2t = 6$ A and $\frac{dI}{dt} = 2$ A/s, so $L\frac{dI}{dt} = 1$ V and $IR = 12$ V.

Step 3: Total:
$1 + 1 + 12 = 14$ V.

Final Answer:
V_AB is 14 V, option (C). \[ \boxed{14\text{ V}} \]