What are the EMF and internal resistance of the source if, at a current of 6A, the power

| February 8, 2021 | education

What are the EMF and internal resistance of the source if, at a current of 6A, the power in the external circuit is 90 W, and at a current of 2A, this power is reduced to 60 W?

I1 = 6 A.

N1 = 90 W.

I2 = 2 A.

N2 = 60 W.

EMF -?

r -?

The current power in the external circuit N is expressed by the formula: N = I2 * R, where I is the current in the external circuit, R is the resistance in the external circuit.

N1 = I12 * R1.

R1 = N1 / I12.

R1 = 90 W / (6 A) 2 = 2.5 Ohm.

N2 = I22 * R2.

R2 = N2 / I22.

Let’s write Ohm’s law for a closed loop for both cases.

I1 = EMF / (R1 + r).

EMF = I1 * (R1 + r).

I2 = EMF / (R2 + r).

EMF = I2 * (R2 + r).

I1 * (R1 + r) = I2 * (R2 + r).

I1 * R1 + I1 * r = I2 * R2 + I2 * r.

I1 * R1 – I2 * R2 = I2 * r – I1 * r.

r = (I1 * R1 – I2 * R2) / (I2 – I1) = (I1 * N1 / I12 – I2 * N2 / I22) / (I2 – I1) = (N1 / I1 – N2 / I2) / (I2 – I1).

r = (90 W / 6 A – 60 W / 2 A) / (2 A – 6 A) = 3.75 Ohm.

EMF = 6 A * (2.5 Ohm + 3.75 Ohm) = 37.5 V.

Answer: r = 3.75 Ohm, EMF = 37.5 V.



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