Question

I have a voltage source of 12V but a light that only burns at 5V. The lamp works on 18 mA. Calculate the resistance that you EXTRA have to include in this series connection.

Answer 1

The resistance that will provide this potential drop is 388.89 ohms.

Step-by-step explanation:

Given;

Voltage source, E = 12 V

Voltage rating of the lamp, V = 5 V

Current through the lamp, I = 18 mA

Extra voltage or potential drop, IR = E- V

IR = 12 V - 5 V = 7 V

The resistance that will provide this potential drop (7 V) is calculated as follows:

IR = V

`R = (V)/(I) = (7 \ V)/(18 * 10^(-3) A) \ = 388.89 \ ohms`

Therefore, the resistance that will provide this potential drop is 388.89 ohms.