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)