Exercises — Define resistance and the ohm

We want $R$, we have $V$ and $I$, so we use the face $R=V/I$. $R=\frac{V}{I}=\frac{6\text{V}}{2\text{A}}=3\Omega$ What we did: divided the push by the flow. Why: that ratio is the definition of resistance.

The unit is the ohm, symbol $\Omega$ (Greek capital omega). One ohm is the resistance when 1 volt drives exactly 1 ampere: $1\Omega =\frac{1\text{V}}{1\text{A}}$

We have $V$ and $R$, we want $I$, so we use $I=V/R$. $I=\frac{9\text{V}}{3\Omega }=3\text{A}$ Why this face: dividing the push by the opposition tells you how much flow gets through.

Missing quantity is $V$, so use $V=IR$. $V=0.25\text{A}\times 48\Omega =12\text{V}$ Why multiply: to recover the push needed, you scale the current by how hard the path fights back. Note the units multiply too: $\text{A}\times \Omega =\text{A}\times \text{V/A}=\text{V}$, confirming the answer really is in volts.

First convert to base units so the formula stays honest: $2.2\text{k}\Omega =2200\Omega$, and $5\text{mA}=0.005\text{A}$. $V=IR=0.005\text{A}\times 2200\Omega =11\text{V}$ Why convert first: the ohm is defined as V/A, so mixing in kilo- and milli- prefixes would silently multiply your answer by the wrong power of ten.

Missing is $I$, so $I=V/R$. $I=\frac{230\text{V}}{575\Omega }=0.4\text{A}$ Why this face: we want flow, and $V/R$ divides the push by the opposition. Units check: $\text{V}/\Omega =\text{V}/(\text{V/A})=\text{A}$ ✔.

On an $I$-vs-$V$ graph the slope is $I/V=1/R$. The marked violet point is $(V,I)=(8\text{V},2\text{A})$. $R=\frac{V}{I}=\frac{8\text{V}}{2\text{A}}=4\Omega$ A steeper line means a larger slope $=1/R$, hence a smaller $R$ — current flows more easily. This is the dual-coding idea from Ohm's Law: steep = easy = low resistance.

Use $I=V/R$. With $R\to 3R$ and $V$ fixed: $I\to \frac{V}{3R}=\frac{1}{3}I$ Current falls to one third. Geometrically: bigger $R$ = smaller slope $1/R$, so the line lies flatter; at the same $V$ the flatter line reaches a lower $I$. The figure below shows both lines and the same-voltage vertical guide.

Use $I=V/R$ and let $R$ approach the extremes. (a) $R\to 0$: $I=V/R\to \infty$. A perfect conductor would let unlimited current flow — this is why an ideal "short circuit" is dangerous; real wires always keep a tiny resistance. (b) $R\to \infty$: $I=V/R\to 0$. An insulator or an open switch passes essentially no current, no matter the voltage. (c) Reversed sign: $I=\frac{-9\text{V}}{3\Omega }=-3\text{A}$. The magnitude is the same 3 A; the minus sign simply says the charge now flows the opposite way round the loop to the direction you first called positive. Why $R$ stays positive: $R=V/I=(-9)/(-3)=+3\Omega$ — reversing the push reverses the flow too, so their ratio is unchanged. Resistance never goes negative for an ordinary resistor. These cases bracket every real situation: zero and infinite $R$ set the current limits, and the sign tells you the direction.

(a) Missing quantity is $V$, and we know $I$ and $R$, so use $V=IR$. Why this face: multiplying the flow by the opposition rebuilds the push that must exist across the resistor. $V=IR=3\text{A}\times 4\Omega =12\text{V}$ (Units: $\text{A}\times \Omega =\text{A}\times \text{V/A}=\text{V}$ ✔.) (b) Power is energy delivered per second. From Power Dissipation in Resistors, $P=VI$, and since $V=IR$ we substitute to get $P=I^{2}R$. Why substitute: it lets us find power straight from the current and resistance without re-using part (a). $P=I^{2}R=(3\text{A}{)}^2\times 4\Omega =9{\text{A}}^2\times 4\Omega =36\text{W}$ Cross-check with $P=VI=12\text{V}\times 3\text{A}=36\text{W}$ ✔ — two routes, same answer, which is how you know the algebra is consistent.

We know $P$ and $V$ but not $I$. Combine $P=VI$ (giving $I=P/V$) with $R=V/I$: $R=\frac{V}{I}=\frac{V}{P/V}=\frac{V^2}{P}=\frac{(240\text{V}{)}^2}{60\text{W}}=\frac{57600}{60}\Omega =960\Omega$ Why $V^2/P$: we eliminated the unknown current by substituting $I=P/V$, leaving only the two things we were given.

In series the current is shared and resistances add (the same charge must pass through both, so the total opposition is the sum): $R_{\text{total}}=3\Omega +5\Omega =8\Omega ,I=\frac{V}{R_{\text{total}}}=\frac{16\text{V}}{8\Omega }=2\text{A}$ Why add: one path forces the same current through both parts, so the pushes needed add up. Now apply $V=IR$ to each with this shared $I=2\text{A}$: $V_3=2\text{A}\times 3\Omega =6\text{V},V_5=2\text{A}\times 5\Omega =10\text{V}$ Check: $6\text{V}+10\text{V}=16\text{V}$ ✔ — the two voltages must add back to the supply.

Why the voltage is shared here: both resistors span the same two nodes, so each has the full $16\text{V}$ across it. Apply $I=V/R$ to each branch separately. $I_3=\frac{16\text{V}}{3\Omega }=5.333\text{A},I_5=\frac{16\text{V}}{5\Omega }=3.2\text{A}$ Why add the branch currents: charge splitting at the node must recombine, so the supply delivers the sum: $I_{\text{total}}=I_3+I_5=5.333+3.2=8.533\text{A}$ Combined resistance from the supply's viewpoint (total push over total flow): $R_{\text{parallel}}=\frac{V}{I_{\text{total}}}=\frac{16\text{V}}{8.533\text{A}}=1.875\Omega$ Sanity check with the reciprocal rule $\frac{1}{R_{\text{p}}}=\frac{1}{3}+\frac{1}{5}=\frac{5+3}{15}=\frac{8}{15}$, so $R_{\text{p}}=\frac{15}{8}=1.875\Omega$ ✔. Note the parallel combination ($1.875\Omega$) is smaller than either resistor — adding a second path always makes current flow easier, unlike series which adds up.

Resistance is defined at any operating point as $R=V/I$, so we can still compute it: $R_1=\frac{2\text{V}}{1\text{A}}=2\Omega ,R_2=\frac{8\text{V}}{2\text{A}}=4\Omega$ The ratio $V/I$ changed (2 → 4 Ω), so this component is non-ohmic: its $I$-vs-$V$ graph is a curve, not a straight line through the origin. Physically, the higher voltage heats the filament, and hotter metal has higher resistance (a temperature effect touched on in Resistivity and Resistance of a Wire). Ohm's law $V\propto I$ only holds when temperature is fixed. (Reversing the supply would flip the sign of both $V$ and $I$ together, so each computed $R$ would still be positive — the curve is symmetric about the origin.)

Start from $1\Omega =1\text{V/A}$. Now expand each:

• volt $=$ joule per coulomb $=\text{J/C}$ (energy per charge, from Voltage and Potential Difference).
• ampere $=$ coulomb per second $=\text{C/s}$ (charge per time, from Electric Current). $1\Omega =\frac{\text{V}}{\text{A}}=\frac{\text{J/C}}{\text{C/s}}=\frac{\text{J}}{\text{C}}\times \frac{\text{s}}{\text{C}}=\frac{\text{J}\cdot \text{s}}{{\text{C}}^2}$ Why this matters: it proves the ohm is a derived unit — nothing new, just a bookkeeping of energy, charge and time.

Resistance: want $R$ from $V$ and target $I$; convert $20\text{mA}=0.02\text{A}$. $R=\frac{V}{I}=\frac{5\text{V}}{0.02\text{A}}=250\Omega$ Power rating: $P=VI=5\text{V}\times 0.02\text{A}=0.1\text{W}$ (equivalently $I^{2}R=(0.02\text{A}{)}^2\times 250\Omega =0.1\text{W}$). A standard $0.25\text{W}$ resistor comfortably survives, giving a safety margin. Why check power: a resistor with the right ohms but too low a power rating will overheat — the correct resistance alone is not enough.