Subelement B: Electrical Math— Topic 9: Ohm’s Law-1
Question 3-9B6
Element 3 (GROL)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?
Explanation
To determine the required series resistance, first calculate the voltage the relay needs to operate. Using Ohm's Law (V = I * R), where I is the current and R is the resistance:
1. **Relay's Operating Voltage (V_relay):**
V_relay = 0.125 A (125 mA) * 500 ohms = 62.5 V
2. **Voltage Drop Needed Across Series Resistor (V_series):**
The total supply voltage is 110 V. The relay only needs 62.5 V. The remaining voltage must be dropped across the series resistor.
V_series = V_supply - V_relay = 110 V - 62.5 V = 47.5 V
3. **Required Series Resistance (R_series):**
Since the resistor is in series with the relay, the current through it will also be 125 mA (0.125 A). Using Ohm's Law again (R = V / I):
R_series = 47.5 V / 0.125 A = 380 ohms
Therefore, a 380-ohm resistor should be connected in series. Options A, B, and D would result in incorrect current levels through the relay, either too high or too low.