When a resistor was connected to the battery, the resistance of 18 ohms was 1 A, and when the resistor

| September 1, 2021 | education

When a resistor was connected to the battery, the resistance of 18 ohms was 1 A, and when the resistor was connected to the battery, the resistance of 8 ohms, the current in the circuit became 1.8 A. Find the EMF and the internal resistance of the battery

R1 = 18 ohms.

I1 = 1 A.

R2 = 8 ohms.

I2 = 1.8 A.

r -?

EMF -?

According to Ohm’s law for a closed circuit, the current in the circuit I is directly proportional to the electromotive force of the EMF current source and is inversely proportional to the sum of the external resistance R and the internal resistance of the current source r: I = EMF / (R + r).

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).

r = (1 A * 18 Ohm – 1.8 A * 8 Ohm) / (1.8 A – 1 A) = 4.5 Ohm.

EMF = 1.8 A * (8 Ohm + 4.5 Ohm) = 22.5 V.

Answer: the current source has EMF = 22.5 V, internal resistance r = 4.5 Ohm.



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