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:

Question

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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 $1 \, \text{V}$.

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:

Solution and Explanation

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