# physics

posted by Anonymous

three resistors with values R1,R2 and R3, respectively are connected in series to a 6-v battery.the total current flowing through the circuit is I=0.6mA.when these three resistors are connected in parallel to a 6v battery,the total current is I=6.2mA. also the current flowing through R2 is I=2mA. determing the individual resistance of the three resistors.note R1<R2<R3

1. Steve

we know that
6/R2 = .002, so R2 = 3KΩ
R1+R2+R3 = 6/.0006 = 10KΩ
So, now we have
R1+R3 = 10K-3K = 7KΩ
6/R1 + 6/R3 = .0062-.002 = .0042

Now just solve for R1 and R3, remembering that R1 < R3

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