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?

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?

• A. 250 ohms
• B. 380 ohms
• C. 500 ohms
• D. 620 ohms

Answer: (B)

To determine the necessary series resistance to operate a relay coil with a resistance of 500 ohms at a current of 125 mA using a 110 V DC source, it's essential first to calculate the total voltage drop across both the relay coil and the series resistor. The first step is to find out the voltage drop across the relay coil. Using Ohm's law (V = I * R), where I is the current in amps and R is the resistance in ohms: - The coil operates at 125 mA, which is 0.125 A. - The resistance of the coil is 500 ohms. Calculating the voltage across the coil: V_coil = I * R_coil = 0.125 A * 500 ohms = 62.5 V Next, we need to determine how much voltage is left for the series resistor when using the 110 V supply: V_resistor = V_supply - V_coil = 110 V - 62.5 V = 47.5 V Now, we can find the value of the series resistor using Ohm's law again. The desired current remains at 0.125 A, so we can rearrange Ohm's law

To determine the necessary series resistance to operate a relay coil with a resistance of 500 ohms at a current of 125 mA using a 110 V DC source, it's essential first to calculate the total voltage drop across both the relay coil and the series resistor.

The first step is to find out the voltage drop across the relay coil. Using Ohm's law (V = I * R), where I is the current in amps and R is the resistance in ohms:

• The coil operates at 125 mA, which is 0.125 A.

• The resistance of the coil is 500 ohms.

Calculating the voltage across the coil:

V_coil = I * R_coil = 0.125 A * 500 ohms = 62.5 V

Next, we need to determine how much voltage is left for the series resistor when using the 110 V supply:

V_resistor = V_supply - V_coil = 110 V - 62.5 V = 47.5 V

Now, we can find the value of the series resistor using Ohm's law again. The desired current remains at 0.125 A, so we can rearrange Ohm's law