In the figure given, for the initial capacitor voltage is zero. The switch is closed at t = 0. The final steady-state voltage across the capacitor is

20 V

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10 V

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5 V

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0 V

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Solution

The correct option is B 10 V
At $(t \to 0^{+})$ , the capacitor act as short-circuit. At $(t \to \infty)$ , the capacitor will become open circuit.

$∴$ Voltage across capacitor

$= \frac{20}{10 + 10} \times 10 = 10 V$

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In the figure given, for the initial capacitor voltage is zero. The switch is closed at t = 0. The final steady-state voltage across the capacitor is

The correct option is B 10 VAt (t→0+) , the capacitor act as short-circuit. At (t→∞), the capacitor will become open circuit. ∴ Voltage across capacitor =2010+10×10=10V

The correct option is B 10 V
At $(t \to 0^{+})$ , the capacitor act as short-circuit. At $(t \to \infty)$ , the capacitor will become open circuit.

$∴$ Voltage across capacitor

$= \frac{20}{10 + 10} \times 10 = 10 V$