A 10 ohm, 20 ohm lamp and 12v battery are connected in a parallel. A current of 1.8 A passes through the battery and a current of 1.2 A passes through the motor.
1) Determine the current through the lamp.
2) Calculate the potential difference across the motor.
3) is this the total resistance of the circuit is greater or smaller than 10 ohms.
2 answers
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R1 = 10 Ohms.
R2 = 20 Ohms.
I1 = 1.2A.
1. I1 + I2 = 1.8A.
1.2 + I2 = 1.8, I2 = 0.6A.
2. V = I1*R1 = 1.2 * 10 = 12 Volts.
3. Rt = R1*R2/(R1+R2) = 10*20/(10+20) = 6.67 Ohms.
R2 = 20 Ohms.
I1 = 1.2A.
1. I1 + I2 = 1.8A.
1.2 + I2 = 1.8, I2 = 0.6A.
2. V = I1*R1 = 1.2 * 10 = 12 Volts.
3. Rt = R1*R2/(R1+R2) = 10*20/(10+20) = 6.67 Ohms.
1 answer