Answer to Question 3

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 A 470 mF capacitor, charged up to 12.0 V is connected to a 100 kW resistor.

(a)    What is the time constant?

(b)   What is the voltage after 10 s?

(c)  How long does it take for the voltage to drop to 2.0 V?          

(a) t = 100 000 W 470 10-6 F = 47 s

(b) V = 12.0 V e-10s/47s = 9.7 V

(c) 2 = 12.0 V e-t/47s ln2 = ln12 + -t/47 t = (0.6931 - 2.485) -47 s = -1.8459 47 s = 87 s