Answer:
The resistance that will provide this potential drop is 388.89 ohms.
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 = 12 V - 5 V = 7 V
The resistance that will provide this potential drop (7 V) is calculated as follows:
[tex]R = \frac{V}{I} = \frac{7 \ V}{18 \times 10^{-3} A} \ = 388.89 \ ohms[/tex]
Therefore, the resistance that will provide this potential drop is 388.89 ohms.
