Question

Asked Apr 1, 2019

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Step 1

**STEP BY STEP METHOD:**

From the given circuit, the switch opens at *t* =0. That is, switch is in closed condition before *t* =0. The capacitor gets open circuited at steady-state condition. Therefore, draw the circuit for *t* =0^{-} (just before open the switch) as shown in Figure 1 and find the value of voltage across the capacitor just before open the switch.

Step 2

From Figure 1, the resistors 2 ohms and 2 ohms are in parallel. The equivalent resistance of two 2 ohms resistors is 1 ohm. Then, the equivalent 1 ohm resistance is connected in series with the 2 ohm resistor. Therefore, the equivalent resistance of 1 ohm and 2 ohm resistor is 3 ohms. Now, the equivalent resistance across the capacitor is 3 ohms.

Consider the analysis from Figure 1, and apply the voltage division rule to find the voltage across capacitor for *t* =0^{-} as follows.

Step 3

As the switch is in open condition for *t* =0, the switch continues to be in open condition for *t* =0+. Therefore, redraw the circuit for *t* ...

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