When a $4 \Omega$ resistor is connected across the terminals of a 12 V battery, the number of coulombs passing through the resistor per second is:(a) 0.3(b) 3(c) 4(d) 12
Given:
Potential difference, V=12 volt
Resistance, R = 4 ohms
To find:
Current, I which is same as coulombs per second
Solution:
According to the ohm's law:
$\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}$
$\mathrm{I}=\frac{\mathrm{12}}{\mathrm{4}}$
= 3 A
The number of coulombs passing through the resistor per second is 3.
Answer is (b)
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