Find the EMF and the internal resistance of the battery if at a current of 15 A it gives

| August 5, 2021 | education

Find the EMF and the internal resistance of the battery if at a current of 15 A it gives 135 W power to the external circuit, and at a current of 6 A the power is 64.8 W

I1 = 15 A.

N1 = 135 W.

I2 = 6 A.

N2 = 64.8 W.

EMF -?

r -?

The current power N in the circuit is determined by the formula: N = I ^ 2 * R, where I is the current in the circuit, R is the external resistance of the circuit.

R = N / I ^ 2.

R1 = N1 / I1 ^ 2.

R1 = 135 W / (15 A) ^ 2 = 0.6 Ohm.

R2 = N2 / I2 ^ 2.

R2 = 64.8 W / (6 A) ^ 2 = 1.8 V.

Let’s write Ohm’s law for a closed loop: I1 = EMF / (R1 + r), I ^ 2 = EMF / (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).

r = (15 A * 0.6 Ohm – 6 A * 1.8 Ohm) / (6 A – 15 A) = 0.2 Ohm.

EMF = I1 * (R1 + r).

EMF = 15 A * (0.6 Ohm + 0.2 Ohm) = 12 V.

Answer: r = 0.2 Ohm, EMF = 12 V.



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