Question

Use Ohm’s law to determine the potential difference across the 3 $\Omega$ resistor in the circuit shown in the following diagram when key is closed:

Answer 1

Total Resistance ($R_{\text{total}}$): All resistors (1 $\Omega$, 2 $\Omega$, 3 $\Omega$) are in series:
$R_{\text{total}} = 1 + 2 + 3 = 6 \, \Omega.$
Current in the Circuit ($I$): Using Ohm’s law ($I = \frac{V}{R_{\text{total}}}$):
$I = \frac{2}{6} = 0.333 \, \text{A}.$
Potential Difference Across the 3 $\Omega$ Resistor ($V = IR$):
$V = 0.333 \times 3 = 1 \, \text{V}.$
The potential difference across the 3 $\Omega$ resistor is $1 \, \text{V}$.