When a 27.0-V emf device is placed across two resistors in series, a current of

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Explanation / Answer

Given that

When a 27.0-V emf device is placed across two resistors in series, a current of 12.0 A is flowing in each of the resistors.

Then R resistance of the circuit is given by ohms law V =IR

Then R =V/I =27/12 =2.25ohms

When the same emf device is placed across the same two resistors in parallel, the current through the emf device is 54.0 A

Again using the ohms law we can find the ressitance of the circuit in parallel then

R =V/I =27/54 =0.5A

The equivalent resistance in series is given by

Req =R1+R2 =2.25 ===> R1 =2.25-R2

The equivalent resistance in parallel is given by

1/Req =1/R1+1/R2

1/0.5 =1/(2.25-R2)+1/R2

Now solve the quadratic equation you will get R2 then we know that

2.25 =R1+R2

R1 =2.25-R2 you will get R1

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