Ohm's Law Practice

Show your work: formula, substitution, then the answer with units.

$$V=IRI=\frac{V}{R}R=\frac{V}{I}$$

1. $V=12\mathrm{V}$, $R=4.0\Omega$. Find $I$.

Answer

$I=\frac{V}{R}=\frac{12\mathrm{V}}{4.0\Omega }=3.0\mathrm{A}$

2. $I=0.50\mathrm{A}$, $R=24\Omega$. Find $V$.

Answer

$V=IR=(0.50\mathrm{A})(24\Omega )=12\mathrm{V}$

3. $V=9.0\mathrm{V}$, $I=0.30\mathrm{A}$. Find $R$.

Answer

$R=\frac{V}{I}=\frac{9.0\mathrm{V}}{0.30\mathrm{A}}=30\Omega$

4. A kettle draws $12.5\mathrm{A}$ at $120\mathrm{V}$. Find its resistance.

Answer

$R=\frac{120\mathrm{V}}{12.5\mathrm{A}}=9.6\Omega$

Low resistance means large current — which is exactly why a kettle needs its own heavy circuit.

5. A $60\Omega$ resistor carries $0.20\mathrm{A}$. Find the potential difference across it.

Answer

$V=(0.20\mathrm{A})(60\Omega )=12\mathrm{V}$

6. Two $10\Omega$ resistors in series across $12\mathrm{V}$. Find the total resistance and the current.

Answer

$R_T=10+10=20\Omega$, so $I=\frac{12\mathrm{V}}{20\Omega }=0.60\mathrm{A}$

7. The same two resistors in parallel across $12\mathrm{V}$.

Answer

$\frac{1}{R_T}=\frac{1}{10}+\frac{1}{10}=\frac{2}{10}$, so $R_T=5.0\Omega$ and $I=\frac{12\mathrm{V}}{5.0\Omega }=2.4\mathrm{A}$

Note that the total resistance is less than either resistor. Adding a second path always makes it easier for current to flow.

8. A phone charger supplies $5.0\mathrm{V}$ at $2.0\mathrm{A}$. Find its power, and the energy delivered in 30 minutes.

Answer

$P=VI=(5.0\mathrm{V})(2.0\mathrm{A})=10\mathrm{W}$

$E=Pt=(10\mathrm{W})(1800\mathrm{s})=18000\mathrm{J}$

Seconds, not minutes — a watt is a joule per second.

Reference: Ohm’s Law, Series and Parallel Circuits