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: Understanding the Figure:
From A to B the circuit has a 4 F capacitor with its positive plate toward A, then a 0.5 H inductor, then a 2 \(\Omega\) resistor. The charge on the positive plate is \(q = t^2 - 5\) coulomb.

Step 2: Current:
The charge on the plate connected to A is increasing, so the current flows from A into the capacitor and on through L and R toward B. Its value is
\[ I = \frac{dq}{dt} = 2t \]
At \(t = 3\) s, \(I = 6\) A, and \(\frac{dI}{dt} = 2\) A/s.

Step 3: Voltage drops from A to B:
Capacitor: \(\frac qC = \frac{3^2 - 5}{4} = \frac44 = 1\) V.
Inductor: \(L\frac{dI}{dt} = 0.5\times2 = 1\) V.
Resistor: \(IR = 6\times2 = 12\) V.

Step 4: Add them:
\[ V_{AB} = 1 + 1 + 12 = 14\text{ V} \]

Step 5: Why the other options are wrong.
8 V, 12 V and 18 V come from leaving out one drop or using \(q = t^2 - 5\) at the wrong time. For example 12 V is the resistor drop alone.

Final Answer:
\(V_{AB} = 14\) V, option (C). \[ \boxed{14\text{ V}} \]