The rheostat is connected to a current source. at first, the current in the circuit was equal to 3A and the voltage

| September 16, 2021 | education

The rheostat is connected to a current source. at first, the current in the circuit was equal to 3A and the voltage on the rheostat-3v when the resistance of the rheostat changed the current strength decreased to 1.5A, and the voltage on the rheostat increased to 4.5V, find the EMF and the internal resistance of the current source

U1 = 3 V.

I1 = 3 A.

U2 = 4.5 V.

I2 = 1.5 A.

r -?

EMF -?

I = EMF / (R + r).

I1 = EMF / (R1 + r).

Let us express the current strength I1 according to Ohm’s law for a section of the circuit: I1 = U1 / R1.

R1 = U1 / I1.

R1 = 3 V / 3 A = 1 ohm.

EMF = I1 * (R1 + r).

I2 = EMF / (R2 + r).

EMF = I2 * (R2 + r).

R2 = U2 / I2.

R2 = 4.5 V / 1.5 A = 3 ohms.

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 = (3 A * 1 Ohm – 1.5 A * 3 Ohm) / (1.5 A – 3 A) = 1 Ohm.

EMF = 3 A * (1 Ohm + 1 Ohm) = 6 V.

Answer: the rheostat is connected to a current source with EMF = 6 V and internal resistance r = 1 Ohm.



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