A battery operating a large flashlight produces a current of 2 amperes in a light bulb that has a resistance of 4 ohms. The voltage of the battery is
| a. |
0.5 volts. |
|
| b. |
2 volts. |
|
| c. |
8 volts. |
|
| d. |
none of the above. |
Solution:
Given:
Current (I) = 2 A
Resistance (R) = 4 Ohm
According to Ohm's Law: V = I R
Thus: V = (2 A)(4 Ohm) = 8 V
Therefore: The voltage of the battery is: (C) 8 Volts
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