Exercises — Calculate electrical power (P = VI, P = I²R)

WHAT we know: $V$ and $I$ directly. WHY this formula: $P=VI$ uses exactly those two — nothing else to compute. $\boxed{P=VI}$

WHAT we know: $I$ and $R$; $V$ is unknown. WHY this formula: $P=I^{2}R$ needs only current and resistance. Using $P=VI$ would force us to first find $V=IR$ — an extra step. $\boxed{P=I^{2}R}$

WHAT we know: $V$ and $R$; $I$ is unknown. WHY this formula: $P=V^2/R$ uses voltage and resistance directly. $\boxed{P=\frac{V^2}{R}}$

WHAT: $V=5\text{V}$, $I=0.9\text{A}$. WHY $P=VI$: both known directly. $P=VI=5\times 0.9=4.5\text{W}$

WHAT: $I=0.030\text{A}$, $R=220\Omega$; voltage unknown, so avoid $VI$. $P=I^{2}R=(0.030{)}^2\times 220=0.0009\times 220=0.198\text{W}$ Reading it: about $0.20\text{W}$ — comfortably under a standard $0.25\text{W}$ rating (see Resistors and Power Ratings).

WHAT: $V=9\text{V}$, $R=18\Omega$; current unknown. $P=\frac{V^2}{R}=\frac{9^2}{18}=\frac{81}{18}=4.5\text{W}$

WHAT: $P=12\text{W}$, $V=6\text{V}$; want $I$. WHY rearrange $P=VI$: solve for the unknown letter. $I=\frac{P}{V}=\frac{12}{6}=2\text{A}$

WHY $I^{2}R$ is the right lens: $R$ is fixed, so power depends on current as $P\propto I^2$. Doubling $I$ multiplies power by $2^2=4$. $P_{\text{new}}=4\times 2\text{W}=8\text{W}$ This is the squared law from the parent note in action — heat grows fast.

WHY $V^2/R$ is the right lens: $V$ is fixed, so $P\propto 1/R$. Halving $R$ doubles $P$. Numerically, if the first element is $12\Omega$: $P_1=12^2/12=12\text{W}$; the second is $6\Omega$: $P_2=12^2/6=24\text{W}$ — indeed twice as much. $P_2=2\times P_1$

WHY $I^{2}R$: current is the shared, fixed quantity; heat then scales linearly with $R$. $\frac{P_B}{P_A}=\frac{I^2{R}_B}{I^2{R}_A}=\frac{R_B}{R_A}=3$ Wire B dissipates 3× the heat. This is exactly why long, thin (high-$R$) wires run hot — see Heat Dissipation and Cooling.

Step 1 — power (WHY $V^2/R$): we hold $V$ and $R$. $P=\frac{V^2}{R}=\frac{12^2}{6}=\frac{144}{6}=24\text{W}$ Step 2 — energy (WHY $E=Pt$): power is energy per second, so multiply by seconds. Convert time: $5\text{min}=5\times 60=300\text{s}$ (watts demand seconds — see Energy and the Joule). $E=Pt=24\times 300=7200\text{J}=7.2\text{kJ}$

(a) Current (WHY rearrange $P=VI$): we have $P$ and $V$. $I=\frac{P}{V}=\frac{1100}{230}\approx 4.78\text{A}$ (b) Resistance (WHY $R=V^2/P$): rearrange $P=V^2/R$ to isolate $R$; uses the two nameplate numbers directly, avoiding the rounded current. $R=\frac{V^2}{P}=\frac{230^2}{1100}=\frac{52900}{1100}\approx 48.1\Omega$

Step 1 — series current (WHY Ohm on the whole loop): total resistance adds in series. $R_{\text{tot}}=R_1+R_2=4+6=10\Omega ,I=\frac{V}{R_{\text{tot}}}=\frac{10}{10}=1\text{A}$ Step 2 — power in $R_2$ (WHY $I^{2}R$): the same current flows through both in series, and we know $R_2$; this avoids finding $R_2$'s voltage separately. $P_2=I^2{R}_2=1^2\times 6=6\text{W}$ Check the trap: using the full $10\text{V}$ in $V^2/R_2$ would be wrong — $R_2$ only gets its share of the voltage ($V_2=I{R}_2=6\text{V}$, and $6^2/6=6\text{W}$ ✓ agrees).

WHY $V^2/R$: we hold $V$ and $R$, and we want the actual dissipated power. $P=\frac{V^2}{R}=\frac{12^2}{470}=\frac{144}{470}\approx 0.306\text{W}$ $0.306\text{W}>0.25\text{W}$, so it is over its rating — it will overheat and likely burn. To be safe you'd need $R>V^2/P_{\text{max}}=144/0.25=576\Omega$, or a higher-rated part (see Resistors and Power Ratings).

Step 1 — current for each (WHY $I=P/V$): the load draws whatever current delivers $1200\text{W}$. $I_a=\frac{1200}{12}=100\text{A},I_b=\frac{1200}{120}=10\text{A}$ Step 2 — wire heat (WHY $I^{2}R$): the wire's own resistance is fixed and carries this current; its waste depends on current, not the load voltage. $P_a=I_a^2{R}_{\text{wire}}=100^2\times 0.5=5000\text{W}$ $P_b=I_b^2{R}_{\text{wire}}=10^2\times 0.5=50\text{W}$ Conclusion: raising the delivery voltage by $10\times$ cut the current by $10\times$, and the $I^2$ law slashed wire loss by $100\times$ ($5000\to 50\text{W}$). This is why the grid uses high voltage — the squared current dominates transmission heat.

Step 1 — current (WHY $I=P/V$): $I=\frac{P}{V}=\frac{3}{5}=0.6\text{A}=600\text{mA}$ Step 2 — runtime (WHY capacity ÷ current): $2000\text{mAh}$ means the battery can supply $2000\text{mA}$ for one hour, or any current×time product equal to that. $t=\frac{2000\text{mAh}}{600\text{mA}}\approx 3.33\text{h}\approx 3\text{h }20\text{min}$ (Idealised: ignores voltage sag and efficiency losses.)

The formula to use when you know current and resistance but not voltage is ==$P=I^{2}R$. If current in a fixed resistor triples, power multiplies by 9== (because $P\propto I^2$). For a fixed-power load, raising delivery voltage 10× cuts wire heat by 100×.