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)?
plz help.....!
I can tell you a) & b)
a) Assuming after 10 min I=Imax then
I1=V/R1 and so the energy is 1/2*L*I1
b) I=V/Req (Req=(R1*R2)/(R1+R2)) So P=I*V
Thnx :) a) n b) are correct ! plz some one for c), d) n e) parts ???
For c) & e):For broken connection of V then Req is R1+R2 so the equation is :
Imax/2=Imax*e^((-Req/L)*t) and Imax is the Imax from before the connection with the V was broken so Imax=V/R where R=(R1*R2)/(R1+R2)