When an external resistor with a resistance R1 = 40 Ohm is connected to the poles of the EMF source

| March 24, 2021 | education

When an external resistor with a resistance R1 = 40 Ohm is connected to the poles of the EMF source, a current of I1 = 1.5 A flows to the circuit, and when an external resistor with a resistance R2 = 29 Ohm is connected, the current is I2 = 2 A. Then the EMF of the source is

R1 = 40 Ohm.
I1 = 1.5 A.
R2 = 29 ohms.
I2 = 2 A.
EMF -?
Let’s write Ohm’s law for a closed circuit: I = EMF / (R + r), where I is the current in the circuit, EMF is the electromotive force of the current source, R is the external resistance of the circuit, r is the internal resistance of the current source.
Let’s write Ohm’s laws for both cases: I1 = EMF / (R1 + r), I2 = EMF / (R2 + r).
EMF = I1 * (R1 + r).
EMF = I2 * (R2 + r).
I1 * (R1 + r) = I2 * (R2 + r).
I1 * R1 + I1 * r = I2 * R2 + I2 r.
I1 * r – I2 * r = I2 * R2 – I1 * R1.
r = (I2 * R2 – I1 * R1) / (I1 – I2).
r = (2 A * 29 Ohm – 1.5 A * 40 Ohm) / (1.5 A – 2 A) = 4 Ohm.
EMF = 1.5 A * (40 Ohm + 4 Ohm) = 66 V.
Answer: EMF of the current source EMF = 66 V.



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