Correct Answers (a) 0.43A (b) I60=0A, I45=I90=0.26A (c) 3.5C (d) I45=0A, I60=I90

Question

Correct Answers

(a) 0.43A

(b) I60=0A, I45=I90=0.26A

(c) 3.5C

(d) I45=0A, I60=I90=0.16A

(e) 36.2s

Problem 5. In the circuit shown below, the switch has been open for a very long time. We then close the switch. (a) Find the current coming from the battery the moment the switch is closed. (b) After the switch has been closed for a very long time, what is the current through each of the resistors? (c) After the switch has been closed for a very long time, how much charge is on the capacitor? (d) We then open the switch. Find the current through each of the resistors the moment the switch is opened. (e) After opening the switch, how long will it take for the capacitor to reach 20% of its charge just before the switch was opened? 450 900 150m 35V 600

Explanation / Answer

When switch is just closed

The capacitor will act as short circuit so current will be

I=35/(Rnet)

Rnet=45+(90*60)/150=81

So current=35/81=0.43

(b) after long time

Capacitor act as open circuit

Current=35/135=0.26

Current across 60 ohm =0 A

Voltage across C=voltage across 90 ohm=23.4 V

Charge=CV=3.5C

If switch is opened

Current across 45=0

Across 90 and 60 we got 23.4/150=0.16

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