A relay coil has 500 ohms resistance and operates on 125 mA. What value of resistance should be connected in series with it to operate from 110 V DC?

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Multiple Choice

A relay coil has 500 ohms resistance and operates on 125 mA. What value of resistance should be connected in series with it to operate from 110 V DC?

To run the relay coil from 110 V DC while limiting current to 125 mA, you need a resistor in series that drops the remaining voltage. The coil has 500 ohms. Desired current is 0.125 A, so the total resistance needed is Rtotal = V / I = 110 / 0.125 = 880 ohms. The series resistor must make up the difference: Rseries = Rtotal − Rcoil = 880 − 500 = 380 ohms. With 0.125 A through the circuit, the power in the series resistor would be P = I^2 R = (0.125)^2 × 380 ≈ 5.9 W, so a resistor rated well above that (e.g., 10 W) is advisable. The coil itself would dissipate Pcoil = I^2 × Rcoil = (0.125)^2 × 500 ≈ 7.8 W. Therefore, the required series resistance is 380 ohms.

To run the relay coil from 110 V DC while limiting current to 125 mA, you need a resistor in series that drops the remaining voltage.

The coil has 500 ohms. Desired current is 0.125 A, so the total resistance needed is Rtotal = V / I = 110 / 0.125 = 880 ohms. The series resistor must make up the difference: Rseries = Rtotal − Rcoil = 880 − 500 = 380 ohms.

With 0.125 A through the circuit, the power in the series resistor would be P = I^2 R = (0.125)^2 × 380 ≈ 5.9 W, so a resistor rated well above that (e.g., 10 W) is advisable. The coil itself would dissipate Pcoil = I^2 × Rcoil = (0.125)^2 × 500 ≈ 7.8 W.

Therefore, the required series resistance is 380 ohms.

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