Question

In: Physics

When a 23.0-V emf device is placed across two resistors in series, a current of 14.0...

When a 23.0-V emf device is placed across two resistors in series, a current of 14.0 A is flowing in each of the resistors. When the same emf device is placed across the same two resistors in parallel, the current through the emf device is 64.0 A. What is the magnitude of the larger of the two resistances?

Solutions

Expert Solution

Resistance of the first resistor = R1

Resistance of the second resistor = R2

Emf of the device = V = 23 V

Equivalent resistance of the circuit when the resistors are placed in series connected = Rs

Rs = R1 + R2

Current through the circuit when the resistors are in series = Is = 14 A

V = IsRs

23 = 14(R1 + R2)

R1 + R2 = 1.643

R1 = 1.643 - R2

Equivalent resistance of the circuit when the resistors are placed in parallel connection = Rp

R1 + R2 = 1.643

Rp = R1R2/1.643

Rp = 0.6086R1R2

Current through the circuit when the resistors are connected in parallel = Ip = 64 A

V = IpRp

23 = (64)(0.6086R1R2)

R1R2 = 0.59

(1.643 - R2)R2 = 0.59

1.643R2 - R22 = 0.59

R22 - 1.643R2 + 0.59 = 0

R2 = 1.113 or 0.53

R1 = 1.643 - R2

Therefore,

R1 = 0.53 or 1.113

Magnitude of the larger resistance = 1.113


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