When the battery is closed to the resistance R1 = 3.00 Ohm, a current flows in the circuit with a force of I1

| June 16, 2021 | education

When the battery is closed to the resistance R1 = 3.00 Ohm, a current flows in the circuit with a force of I1 = 5.00 A. When connected in parallel to the first resistance of the second resistor R2 = 6.00 Ohm, a current with a force of I2 = 8.00 A flows in an unbranched circuit. Determine the internal resistance of the battery.

R1 = 3 ohms.

I1 = 5 A.

R2 = 6 ohms.

I2 = 8 A.

r -?

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

I1 = EMF / (R1 + r).

When connecting the second resistance R2 in parallel to the first R1, the total resistance will be determined by the formula: R = R1 * R2 / (R1 + R2).

R = 3 ohms * 6 ohms / (3 ohms + 6 ohms) = 2 ohms.

I2 = EMF / (R + r).

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

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

R1 * I1 – R * I2 = r * I2 – r * I1.

r = (R1 * I1 – R * I2) / (I2 – I1).

r = (3 Ohm * 5 A – 2 Ohm * 8 A) / (8 A – 5 A) = – 0.33 Ohm.

An error was made in the problem statement, since the internal resistance cannot be negative.



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