Question

In the given circuit, \(V=10\) V, \(R=10\Omega\), \(L=1\) H, \(C=10\mu F\) and \(V_C(0)=0\). Find \(\dfrac{di(0^+)}{dt}\)

• A. \(0\ \text{A/s}\)
• B. \(10\ \text{A/s}\)
• C. \(-100\ \text{A/s}\)
• D. \(1\ \text{A/s}\)

Hint:

At switching instant:
• Capacitor voltage cannot change suddenly
• Inductor current cannot change suddenly For inductor: \[ V_L=L\frac{di}{dt} \]

Answer 1

The Correct Option is B

Concept: For an inductor: :contentReference[oaicite:4]{index=4} At the instant immediately after switching: \[ t=0^+ \] Important properties:
• Inductor current cannot change instantaneously.
• Capacitor voltage cannot change instantaneously. These initial conditions help determine: \[ \frac{di(0^+)}{dt} \]

Step 1: Write the given values. Given: \[ V=10\ \text{V} \] \[ R=10\Omega \] \[ L=1\ \text{H} \] \[ C=10\mu F \] \[ V_C(0)=0 \] We need: \[ \frac{di(0^+)}{dt} \]

Step 2: Understand the initial condition at \(t=0^+\). Initially capacitor is uncharged: \[ V_C(0)=0 \] Since capacitor voltage cannot change suddenly: \[ V_C(0^+)=0 \] Therefore at \(t=0^+\):
• Capacitor behaves like a short circuit. Also initially: \[ i(0^+)=0 \] Hence voltage across resistor: \[ V_R=iR=0 \]

Step 3: Apply KVL around the loop. Applying Kirchhoff’s Voltage Law: \[ V=V_R+V_L+V_C \] At \(t=0^+\): \[ 10=0+V_L+0 \] Thus: \[ V_L=10\ \text{V} \]

Step 4: Use inductor equation. Using: \[ V_L=L\frac{di}{dt} \] Substitute values: \[ 10=1\cdot \frac{di(0^+)}{dt} \] Hence: \[ \frac{di(0^+)}{dt}=10\ \text{A/s} \]

Step 5: Write final answer. Therefore: \[ \boxed{\frac{di(0^+)}{dt}=10\ \text{A/s}} \] Hence correct option is: \[ \boxed{(B)} \]