**Part A**

The switch has been closed for long, so the capacitor behaves as an open circuit. No current is flowing from the capacitor. Equivalent resistance for the series connection:

R_{eq} = 60 + 40 = 100 Ω

Current, i = V/R = 100V/100 = 1.0 A

The voltage across the 40Ω - resistor:

The switch in the figure has been closed for a very long time.

Part A

What is the charge on the capacitor?

Part B

The switch is opened at . t= 0 s At what time has the charge on the capacitor decreased to 29% of its initial value?

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