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September 3, 2015

Homework Help: Physics

Posted by abdul rahman on Tuesday, April 9, 2013 at 2:44pm.

A circuit consists of a self inductor of L=0.003 H in series with a resistor R1=5 Ohm. Parallel to these is a resistor R2=10 Ohm. A battery of V0=9 volt is driving the circuit.

Current has been running for 10 minutes.

(a) How much energy (in Joules) is now stored in the self-inductor.

(b) How much power (in Watt) is then generated by the battery into the circuit? (ignore internal resistance of the battery).

The connection to the battery is now broken (so that the battery is not connected to the circuit anymore).

(c) How long will it take (in seconds) for the current through R1 to be reduced by 50%?

(d) How long does it take (in seconds) till the energy stored in the self-inductor has been reduced by 50%?

(e) How long will it take (in seconds) for the current in R2 to be reduced by 50% (compared to the highest value right after the battery is disconnected)?

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