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Current and drift, resistivity, power, emf and internal resistance, maximum power transfer, and Kirchhoff's rules for any network.
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By the end of this lesson you will be able to analyze any circuit of batteries and resistors with Kirchhoff's rules and account for its power.
From Physics 2 you know Ohm's law and have combined resistors. From the last lessons you know potential as energy per charge. This lesson connects current to the motion of charges inside conductors, and gives a complete method, Kirchhoff's rules, for any circuit of batteries and resistors, including those that cannot be reduced by series and parallel steps.
| Term | What it means |
|---|---|
| Current | $I = dQ/dt$, the rate charge passes a point, in amperes (C/s). |
| Current density | $J = I/A = nqv_d$, current per unit cross-sectional area. |
| Resistivity | $\rho$, a material's resistance to current, in Ω·m; copper's is $1.7 \times 10^{-8}$. |
| Resistance | $R = V/I = \rho L/A$ for a uniform conductor, in ohms. |
| Emf | $\mathcal{E}$, the work per charge a battery does moving charge through itself. |
| Internal resistance | $r$, the resistance inside a battery that lowers its terminal voltage. |
| Kirchhoff's rules | Currents into a junction sum to zero; voltage changes around a loop sum to zero. |
Current is the rate at which charge crosses a surface: $I = dQ/dt$. In a metal, a field $E$ drives electrons with a small drift velocity $v_d$, and $I = nqv_dA$. The field needed is proportional to the current density, $E = \rho J$, so for a uniform wire $V = EL = \rho LI/A$, which is Ohm's law:
$$V = IR, \qquad R = \frac{\rho L}{A}.$$
A resistor turns electrical energy into heat at the rate $P = IV = I^2R = V^2/R$. A battery supplies an emf $\mathcal{E}$, but its internal resistance $r$ makes its terminal voltage $\mathcal{E} - Ir$. Any network obeys
Another way: picture
Picture a closed loop of pipe full of water, with a pump and a narrow section. The pump raises the pressure; the water loses that pressure pushing through the narrow section. The same amount of water passes every point each second; none is used up. What is used up is the pump's energy, turned to heat by friction in the narrow section. A battery, a resistor and current work the same way.
Another way: steps
In a copper wire there are about $8.5 \times 10^{28}$ free electrons per cubic meter. A current of $1$ A through a wire of $1$ mm² cross section needs a drift speed $v_d = I/nqA$ of only about $0.07$ mm/s. The electrons wander randomly at a million meters per second, but their average motion along the wire is a slow crawl.
A light comes on instantly anyway because the field that drives the electrons spreads through the circuit at nearly the speed of light, starting every electron moving at once, like water in a full hose. By convention, current points the way positive charge would move, opposite to the electrons' drift.
Resistivity is a property of the material; resistance, of the object. A longer wire has more material to push through, so $R \propto L$; a thicker wire has more paths side by side, so $R \propto 1/A$. Copper and aluminum, with $\rho$ near $10^{-8}$ Ω·m, make wires; nichrome, $100$ times higher, makes heating elements; glass and rubber, $10^{12}$ or more, insulate.
Resistivity of metals rises with temperature, by about $0.4$ percent per degree for copper, because hotter atoms scatter electrons more. A tungsten filament's resistance hot is about ten times its value cold. Ohm's law, $V \propto I$, holds for metals at fixed temperature, but not for diodes, filaments or many other devices.
A charge $dq$ falling through a potential difference $V$ gives up energy $V\,dq$, so the power is $P = V\,dq/dt = IV$. In a resistor this becomes heat: $P = I^2R = V^2/R$. Which form to use depends on what is fixed. In series, where current is shared, the larger resistor takes more power. In parallel, where voltage is shared, the smaller resistor takes more.
Utilities bill for energy in kilowatt-hours: one kilowatt for one hour, $3.6$ MJ. A $1500$ W heater running eight hours uses $12$ kWh, about two dollars at typical American rates.
A battery's emf is the energy it gives each coulomb passing through it. Its chemistry has resistance, so under load some of that energy is lost inside, and the terminal voltage is $V = \mathcal{E} - Ir$. A car battery with $\mathcal{E} = 12.6$ V and $r = 0.010$ Ω delivering $200$ A to the starter drops to $10.6$ V.
The power delivered to a load $R$ is $\mathcal{E}^2R/(R + r)^2$, which is greatest when $R = r$: the maximum power transfer condition. At that point half the energy is wasted inside the source, so power systems avoid it, but audio amplifiers and radio antennas are designed to match their loads to get the most signal through.
The junction rule is charge conservation: charge cannot pile up at a junction in a steady circuit. The loop rule is energy conservation: a charge going around a loop and returning to its start must end with the same potential energy, so the gains through batteries equal the drops across resistors.
Sign conventions matter. Crossing a resistor in the direction of the assumed current, the potential drops by $IR$. Crossing a battery from $-$ to $+$, it rises by $\mathcal{E}$. With $n$ unknown currents, write junction equations at all but one junction and loop equations for the rest; any extra equations will be combinations of these. A node-potential approach, with one unknown potential per junction, often needs fewer equations.
An ammeter measures current, so it goes in series with the branch and should have very small resistance, or it would change the current it measures. A voltmeter measures potential difference, so it goes in parallel across a component and should have very large resistance, or it would draw current and change the voltage.
Digital multimeters have voltmeter resistances of about $10$ MΩ, plenty for most circuits, but they can disturb measurements in high-resistance circuits such as those in sensors and nerve cells. Knowing how a meter loads a circuit is part of reading it correctly. The same reasoning explains why an ammeter placed across a battery by mistake is dangerous: its tiny resistance makes it a near short circuit, and the large current can blow its fuse or damage the battery. Checking where a meter's leads go before switching on is a habit worth building early, in the lab and at home.
Checking an answer. Power supplied by the batteries must equal power dissipated in the resistors. Currents at every junction must balance. The potential found by going around two different paths between the same points must agree. And a series combination must exceed its largest resistor; a parallel one must be less than its smallest.
Kirchhoff's rules assume a steady state: currents not changing, so no charge accumulating anywhere and no changing magnetic fields inducing extra voltages. In the next lessons, with capacitors charging and inductors present, the loop rule is kept by including the capacitor's $q/C$ and the inductor's $L\,dI/dt$ as voltage changes.
The series and parallel formulas follow from the rules. In series, one current and voltages adding give $R_1 + R_2$. In parallel, one voltage and currents adding give $1/R = 1/R_1 + 1/R_2$. When a network cannot be broken into such groups, as with a bridge circuit, the rules still apply directly.
American homes receive $120$ V on most outlets, and $240$ V for dryers, ovens and heat pumps. Every outlet on a circuit is in parallel, so each device gets the full voltage and turning one off does not affect the others. The currents add, and a circuit breaker, rated at $15$ or $20$ A, opens if the total grows large enough to overheat the wires.
That is why running a space heater and a hair dryer on one circuit trips the breaker: $12.5$ A plus $12.5$ A exceeds $20$ A. Wires are sized by gauge so that their resistance, and the $I^2R$ heating in them, stays small at the rated current.
A $1500$ W space heater draws $12.5$ A from a $120$ V outlet, most of what a $15$ A circuit can carry. In cold states like Minnesota, fire departments warn against plugging heaters into extension cords, whose thin wires have enough resistance to heat up at that current, and against sharing a circuit with other large loads.
Circuit breakers protect the wiring, not the appliance. A breaker senses the total current on its circuit and opens when it stays above the rating, before the wires in the walls get hot enough to damage insulation. Ground-fault breakers also compare current out and back, and trip if a few milliamperes leak through a person.
Power plants deliver electricity at hundreds of kilovolts. The reason is $I^2R$: for a given power, raising the voltage lowers the current, and the heating in the lines falls as the square. Sending $1000$ MW at $500$ kV needs $2000$ A; at $50$ kV it would need $20{,}000$ A and lose a hundred times as much power in the same wires.
The high-voltage lines that cross the country, run by utilities and regional grid operators, lose only a few percent of the power they carry. Transformers, which later lessons explain, step the voltage down in stages to the $120$ V delivered to homes.
It seems natural that current weakens as it passes through resistors, so that less returns to the battery than left it. But charge is conserved: the current out of a resistor equals the current in. What the resistor takes is energy, so the potential drops across it, not the current.
A related error is to think a battery supplies a fixed current. A battery supplies a roughly fixed emf; the current depends on the circuit connected to it, and a battery shorted by a wire drives a large current limited only by its internal resistance.
A $4.0$ Ω resistor is in series with a parallel pair of $6.0$ Ω and $12$ Ω across $24$ V. Combine the pair.
$R_{\parallel} = \dfrac{6.0 \times 12}{18} = 4.0\ \Omega$
Parallel rule.
Find the total resistance.
$R = 4.0 + 4.0 = 8.0\ \Omega$
Series.
Find the battery current.
$I = \dfrac{24}{8.0} = 3.0\ \text{A}$
Ohm's law.
Find the voltage across the pair.
$V = 3.0 \times 4.0 = 12\ \text{V}$
The other $12$ V drops across the series resistor.
Find the current in each branch.
$I_6 = \dfrac{12}{6.0} = 2.0\ \text{A}, \quad I_{12} = \dfrac{12}{12} = 1.0\ \text{A}$
They add to $3.0$ A.
A $9.0$ V battery with $r = 0.50$ Ω drives a $4.0$ Ω lamp. Find the current.
$I = \dfrac{9.0}{4.5} = 2.0\ \text{A}$
Internal and load resistances in series.
Find the terminal voltage.
$V = 9.0 - 2.0 \times 0.50 = 8.0\ \text{V}$
The internal drop is $1.0$ V.
Find the lamp's power.
$P = 2.0^2 \times 4.0 = 16\ \text{W}$
$I^2R$.
Find the power wasted inside.
$P_r = 2.0^2 \times 0.50 = 2.0\ \text{W}$
It warms the battery.
Check the energy balance.
$\mathcal{E}I = 9.0 \times 2.0 = 18\ \text{W} = 16 + 2.0$
Power in equals power out.
Find the load that takes the most power.
$R = r = 0.50\ \Omega: \ P = \dfrac{81 \times 0.50}{1.0} = 40.5\ \text{W}$
Matched load.
A $12$ V battery with a $2.0$ Ω resistor and a $6.0$ V battery with a $3.0$ Ω resistor both feed a $6.0$ Ω resistor, positive terminals up. Write the branch currents in terms of the top node potential $V$.
$I_1 = \dfrac{12 - V}{2.0}, \quad I_2 = \dfrac{6.0 - V}{3.0}, \quad I_3 = \dfrac{V}{6.0}$
One unknown instead of three.
Apply the junction rule.
$\dfrac{12 - V}{2.0} + \dfrac{6.0 - V}{3.0} = \dfrac{V}{6.0}$
Current in equals current out.
Multiply through by six.
$36 - 3V + 12 - 2V = V$
Clear the fractions.
Solve for the node potential.
$V = 8.0\ \text{V}$
$48 = 6V$.
Find each current.
$I_1 = 2.0\ \text{A}, \quad I_2 = -0.67\ \text{A}, \quad I_3 = 1.33\ \text{A}$
The negative sign means the $6.0$ V battery is being charged.
Check the junction rule.
$2.0 - 0.67 = 1.33$
It balances.
Check the power balance.
$12 \times 2.0 - 6.0 \times 0.67 = 20 = 2.0^2(2.0) + 0.67^2(3.0) + 1.33^2(6.0)$
Supplied equals dissipated plus stored in the charged battery.
Combine the parallel pair.
$R_{\parallel} = 3.0\ \Omega$
Two equal resistors in parallel give half.
Add the series resistor.
$R = 3.0 + 6.0 = 9.0\ \Omega$
Series.
Divide the voltage by the total.
A copper wire has resistance $9$ Ω. It is drawn out to twice its length, keeping the same volume of copper. What is its resistance now?
Complete the worked solution: resistors of $30$ Ω and $20$ Ω in parallel are connected in series with a $3$ Ω resistor across a $75$ V battery. Find the parallel pair's resistance, the circuit's total resistance, both in Ω, and the current from the battery, in A.
Combine the parallel pair.
$R_{\parallel} = \dfrac{R_2R_3}{R_2 + R_3} =$ p
Same voltage across both; currents add.
Add the series resistor.
$R_{\text{total}} = R_1 + R_{\parallel} =$ t
Same current through both.
Divide the battery voltage by the total.
$I = \dfrac{V}{R_{\text{total}}} =$ c
Ohm's law for the whole circuit.
Check the split in the parallel pair.
$I_2 : I_3 = R_3 : R_2$
More current takes the easier path.
Match each quantity to its expression.
| $R_1 + R_2$ | $R_1R_2/(R_1 + R_2)$ | $\rho L/A$ | $I^2R$ | |
|---|---|---|---|---|
| in series | ||||
| in parallel | ||||
| a uniform wire | ||||
| power in a resistor |
A battery with emf $30$ V and internal resistance $2$ Ω drives a $8$ Ω load. Fill in the current in A, the voltage across the battery's terminals in V, and the power delivered to the load in W.
| value | |
|---|---|
| current (A) | |
| terminal voltage (V) | |
| power to load (W) |
A $8$ V battery with internal resistance $3$ Ω is connected to a variable load resistance $R$ (Ω). Write the power delivered to the load, in W, as a formula in $R$.
Answer:
A $8$ V battery in series with a $1.0$ Ω resistor and a $3$ V battery in series with a $2.0$ Ω resistor are both connected, positive terminals toward the top, across a $2.0$ Ω resistor. What current flows through that $2.0$ Ω middle resistor, in A?
Answer: A through the middle resistor
A homeowner in Duluth, Minnesota, plugs a $1500$ W space heater into a $120$ V outlet. What is the resistance of its heating element while it runs, in ohms?
Answer: Ω for the heating element
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A $10$ V battery with internal resistance $2$ Ω is connected to a variable load resistance $R$ (Ω). Write the power delivered to the load, in W, as a formula in $R$.
Answer:
You can solve DC circuits. Explain to someone why the current leaving a resistor is the same as the current entering it.
20. Your turn: three $6.0$ Ω resistors, two in parallel and that pair in series with the third, across $18$ V. Find the battery current., step 3
$I = \dfrac{18}{9.0} = 2.0\ \text{A}$
Ohm's law.