Part 3 · Chapter 27

Circuits

Last chapter gave us current and resistance one element at a time. Now we wire those elements together into working circuits. The new ingredient is a device that does work on charge — an emf source such as a battery — that keeps the current flowing. This chapter builds the tools to predict the current and the potential at any point in a circuit: the single-loop result \(i = \mathcal{E}/(R+r)\), the rules for combining resistors in series and in parallel, Kirchhoff's two rules for any tangled multiloop network, and finally the slow, exponential charging and discharging of a capacitor through a resistor.

Fundamentals of Physics Prof. Mithun Mondal Reading time ≈ 60 min
i What you'll learn
  • Electromotive force as the work per unit charge an emf device supplies, \(\mathcal{E} = \dfrac{dW}{dq}\), measured in volts — not a force at all.
  • The single-loop circuit with a real battery of internal resistance \(r\), giving \(i = \dfrac{\mathcal{E}}{R + r}\), and the terminal voltage \(V = \mathcal{E} - ir\).
  • Combining resistors: in series \(R_{\text{eq}} = \sum R_{j}\), in parallel \(\dfrac{1}{R_{\text{eq}}} = \sum \dfrac{1}{R_{j}}\).
  • Kirchhoff's rules for any network — the junction (current) rule \(\sum i_{\text{in}} = \sum i_{\text{out}}\) and the loop (voltage) rule \(\sum V = 0\) around any closed loop.
  • RC circuits: a capacitor charges as \(q = C\mathcal{E}\big(1 - e^{-t/RC}\big)\) and discharges as \(q = q_{0}\,e^{-t/RC}\), set by the time constant \(\tau = RC\).
Section 27-1

What Is Physics?

In the last chapter we learned what a current is and how a single conductor resists it. But a lone resistor does nothing on its own; to be useful, current must be made to flow round and round through a connected set of elements — a circuit. Flip a switch, start a car, or charge a phone, and you set a circuit working. The physics question of this chapter is squarely practical: given a battery and an arrangement of resistors and capacitors, what current flows in each branch, and what is the potential at each point?

The crucial new idea is that a steady current cannot keep itself going. Left alone, the field inside a wire pushes charge "downhill" in potential until the flow stops. To maintain a current we need a device that does work to lift charge back "uphill" — a charge pump. That device, and the energy bookkeeping around the loop it drives, is where we begin.

Section 27-2

"Pumping" Charges — Electromotive Force

To drive a steady current we need an emf device — a battery, an electric generator, a solar cell, a fuel cell. Such a device does work on charge carriers, raising them from its low-potential terminal to its high-potential terminal, against the electric field that would otherwise drive them the other way. The measure of this pumping ability is the electromotive force, or emf, \(\mathcal{E}\).

Definition of emf
\[ \mathcal{E} = \frac{dW}{dq} \]
The emf is the work \(dW\) the device does per unit charge \(dq\) it moves from its negative to its positive terminal. Its SI unit is the volt — the same as potential difference — so despite the name, an emf is an energy-per-charge, not a force.
"Electromotive force" is a historical misnomer. It is measured in volts, carries units of joules per coulomb, and is not a force in newtons. Think of an emf device as a pump that does a fixed amount of work on every coulomb it pushes through itself — a water pump raising water to a height, against gravity, so it can flow back down through the plumbing.
Section 27-3

Work, Energy, and EMF

Follow a charge \(dq\) once around a circuit driven by an ideal emf device (one with no internal resistance). Inside the device, chemical or other forces do work \(dW = \mathcal{E}\,dq\) lifting the charge to the high-potential terminal. Out in the external circuit, that same charge gives up its energy as it falls through the resistors. Over a full loop the energy supplied equals the energy dissipated — energy is conserved.

Power delivered by an emf device
\[ P = i\,\mathcal{E} \]
In time \(dt\) a charge \(dq = i\,dt\) passes through, and the device does work \(dW = \mathcal{E}\,dq = \mathcal{E}\,i\,dt\). The rate is the power the source feeds into the circuit. For an ideal source all of it reaches the external elements; for a real source, some is lost inside.
Section 27-4

Calculating the Current in a Single-Loop Circuit

Take the simplest complete circuit: an ideal emf \(\mathcal{E}\) driving a single resistance \(R\). There are two standard ways to find the current, and both must agree.

The energy method equates the power delivered by the source to the power dissipated in the resistor: \(i\mathcal{E} = i^{2}R\), which at once gives \(i = \mathcal{E}/R\). The potential method walks once around the loop, adding up the changes in potential and setting their sum to zero (we return to where we started). Starting at the low terminal and going with the current: a rise \(+\mathcal{E}\) across the battery, then a drop \(-iR\) across the resistor.

Single ideal loop
\[ \mathcal{E} - iR = 0 \qquad\Longrightarrow\qquad i = \frac{\mathcal{E}}{R} \]
This balancing of potential changes around a loop is the seed of Kirchhoff's loop rule. The two methods are not rivals — energy conservation and the single-valuedness of potential are two faces of the same physics.
Section 27-5

Internal Resistance and Resistances in Series

A real battery is not ideal: its own materials resist the current passing through them. We model this with an internal resistance \(r\) in series with an ideal emf. Now the loop equation picks up a second drop, this one inside the source:

Real battery driving a resistor
\[ \mathcal{E} - ir - iR = 0 \qquad\Longrightarrow\qquad i = \frac{\mathcal{E}}{R + r} \]
The internal resistance simply adds to the external resistance. As \(r \to 0\) the battery approaches ideal behaviour. The voltage you would actually measure across the battery's terminals is the terminal voltage \(V = \mathcal{E} - ir\), always less than \(\mathcal{E}\) while current flows.

Resistors connected one after another, so the same current passes through each in turn, are in series. The potential drops add, so the chain behaves like a single resistor whose resistance is their sum.

Resistances in series
\[ R_{\text{eq}} = R_{1} + R_{2} + \cdots = \sum_{j} R_{j} \]
Series elements share one current and split the voltage. The equivalent resistance is larger than the largest individual resistor — adding more resistance in line always impedes the flow more.
Section 27-6

Potential Difference Between Two Points

Once the current is known, the potential difference between any two points follows by walking between them and tallying the rises and drops along the way — the answer is independent of the path you take. For the real battery above, the difference between the terminals \(a\) (high) and \(b\) (low), measured through the battery, is

Terminal voltage of a real battery
\[ V_{a} - V_{b} = \mathcal{E} - ir \]
Going the other way, through the external resistor, gives the same value as \(iR\) — they must agree because potential is single-valued. When the battery is charging (current forced backward through it) the sign of the \(ir\) term flips and the terminal voltage exceeds the emf.
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A rule for walking a circuit
Resistor: move with the current and the potential drops by \(iR\); move against it and the potential rises by \(iR\).

EMF device: move from the \(-\) terminal to the \(+\) terminal and the potential rises by \(\mathcal{E}\) (whatever the current direction). Apply these two sign rules consistently and any potential difference falls out by simple addition.

Section 27-7

Multiloop Circuits and Kirchhoff's Rules

Most real circuits cannot be reduced to a single loop. They branch at junctions and contain several independent loops, often with more than one battery. Two conservation principles — first written down by Gustav Kirchhoff — handle any such network.

Kirchhoff's junction (current) rule
\[ \sum i_{\text{in}} = \sum i_{\text{out}} \]
At any junction, charge cannot pile up in the steady state, so the current flowing in equals the current flowing out. This is conservation of charge.
Kirchhoff's loop (voltage) rule
\[ \sum_{\text{loop}} V = 0 \]
The algebraic sum of the changes in potential around any closed loop is zero — you return to your starting potential. This is conservation of energy.

The recipe is mechanical. Label an assumed direction for the current in every branch (a wrong guess merely returns a negative answer). Write the junction rule at the junctions, and the loop rule around enough independent loops, until you have as many equations as unknown currents. Then solve. The single-loop result of Section 27-5 is just the simplest special case.

Section 27-8

Resistances in Parallel

Resistors connected across the same two points, so each feels the same voltage, are in parallel. The currents through them add, and the combination conducts more easily than any one branch alone, so the reciprocals of the resistances add.

Resistances in parallel
\[ \frac{1}{R_{\text{eq}}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \cdots = \sum_{j}\frac{1}{R_{j}} \]
Parallel elements share one voltage and split the current. The equivalent resistance is smaller than the smallest individual resistor — adding another path always makes it easier for charge to get across.
Series adds resistance; parallel adds conductance. The mirror image is exact and worth memorising alongside the capacitor rules of Chapter 25 — where it is the other way around. Resistors in series, like capacitors in parallel, simply add. Resistors in parallel, like capacitors in series, add as reciprocals. Mixing these up is the single most common circuit-analysis error.
Section 27-9

The Ammeter and the Voltmeter

An ammeter measures current and must be placed in series in the branch of interest, so the whole current passes through it. To avoid disturbing that current, an ideal ammeter has negligible resistance. A voltmeter measures potential difference and is connected in parallel across the two points of interest. To avoid diverting current from the circuit, an ideal voltmeter has very large resistance.

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Connect them correctly
Ammeter in series, low resistance; voltmeter in parallel, high resistance.

A real meter is imperfect — a real ammeter has small but nonzero resistance, a real voltmeter draws a tiny current — so each slightly perturbs the very quantity it reads. The art of good measurement is keeping that perturbation negligible compared with the circuit's own resistances.

Section 27-10

RC Circuits

So far the currents have been steady. Add a capacitor and the current becomes time-dependent. Consider a resistor \(R\) and capacitor \(C\) in series with an emf. Applying the loop rule with \(V_{C} = q/C\) and \(i = dq/dt\) gives a first-order differential equation, \(\mathcal{E} = iR + q/C\), whose solution describes the capacitor charging.

Charging a capacitor
\[ q(t) = C\mathcal{E}\left(1 - e^{-t/RC}\right), \qquad i(t) = \frac{\mathcal{E}}{R}\,e^{-t/RC} \]
The charge climbs toward its final value \(C\mathcal{E}\) while the current starts at \(\mathcal{E}/R\) and decays to zero. At the first instant the uncharged capacitor acts like a plain wire; long after, the fully charged capacitor acts like a break in the circuit.

If instead a charged capacitor is allowed to drive current through the resistor with no emf present, it discharges: the loop rule \(q/C + R\,dq/dt = 0\) gives a pure exponential decay.

Discharging a capacitor
\[ q(t) = q_{0}\,e^{-t/RC}, \qquad \tau = RC \]
The product \(\tau = RC\) is the time constant, with units of seconds. In one time constant the charge falls to \(1/e \approx 37\%\) of its initial value; in charging, it rises to \(1 - 1/e \approx 63\%\) of its final value. A large \(R\) or \(C\) makes the process slow.
Why the lag matters. The exponential timing of an RC circuit is not a nuisance — it is the basis of windshield-wiper delays, camera-flash recharge, the blink rate of turn signals, and the timing elements inside countless electronic devices. Choose \(R\) and \(C\), and you choose how long the circuit waits.
Worked Examples

Putting It to Work

1 Current and terminal voltage of a real battery

Problem. A battery of emf 12 V and internal resistance 0.50 Ω drives a 5.5 Ω resistor. Find the current and the battery's terminal voltage.

Solution. Use \(i = \mathcal{E}/(R + r)\), then \(V = \mathcal{E} - ir\).

i = E/(R+r); V = E − ir
\[ i = \frac{12}{5.5 + 0.50} = 2.0\,\mathrm{A}, \qquad V = 12 - (2.0)(0.50) = 11\,\mathrm{V} \]

The terminal voltage falls a full volt short of the 12 V emf — the missing volt is dropped across the internal resistance and never reaches the external circuit.

2 Series versus parallel equivalent

Problem. Three resistors of 4.0 Ω, 6.0 Ω, and 12 Ω are combined. Find the equivalent resistance (a) all in series and (b) all in parallel.

Solution. Series resistances add; parallel reciprocals add.

Series and parallel combination
\[ R_{\text{series}} = 4.0 + 6.0 + 12 = 22\,\Omega, \qquad \frac{1}{R_{\text{par}}} = \frac{1}{4.0}+\frac{1}{6.0}+\frac{1}{12} = \frac{1}{2.0}\;\Rightarrow\; R_{\text{par}} = 2.0\,\Omega \]

The series value tops the largest resistor; the parallel value sits below the smallest. The same three parts, wired two ways, differ by a factor of eleven.

3 Power accounting in a battery

Problem. A 12 V battery of internal resistance 0.040 Ω drives a 0.20 Ω external load, giving a current of 50 A. Find the rate the emf supplies energy, the rate dissipated inside the battery, and the rate delivered to the load.

Solution. Use \(P_{\mathcal{E}} = i\mathcal{E}\), \(P_{r} = i^{2}r\), and \(P_{R} = i^{2}R\).

P = iE, i²r, i²R
\[ P_{\mathcal{E}} = (50)(12) = 600\,\mathrm{W}, \quad P_{r} = (50)^{2}(0.040) = 100\,\mathrm{W}, \quad P_{R} = (50)^{2}(0.20) = 500\,\mathrm{W} \]

The books balance: \(600 = 100 + 500\). One-sixth of the chemical energy is wasted heating the battery itself — exactly why a high-current load makes a battery warm.

4 Charging time of an RC circuit

Problem. An emf of 10 V charges a 1.0 µF capacitor through a 2000 Ω resistor. Find the time constant, the final charge, and the time to reach half the final charge.

Solution. Compute \(\tau = RC\) and \(Q = C\mathcal{E}\); then solve \(q = Q(1 - e^{-t/\tau})\) for \(q = Q/2\).

τ = RC; Q = CE; t = τ ln2
\[ \tau = (2000)(1.0\times10^{-6}) = 2.0\,\mathrm{ms}, \quad Q = 10\,\mu\mathrm{C}, \quad t_{1/2} = \tau\ln 2 \approx 1.4\,\mathrm{ms} \]

Reaching halfway takes about 0.69 time constants — and the capacitor is over 99% charged after only about five time constants, roughly 10 ms here.

5 Discharging a capacitor

Problem. The capacitor of Example 4, charged to 10 V, is disconnected from the emf and allowed to discharge through the same 2000 Ω resistor. What is the voltage across it 5.0 ms later?

Solution. Voltage decays as \(V = V_{0}\,e^{-t/\tau}\) with \(\tau = 2.0\,\mathrm{ms}\).

V = V₀ e^(−t/τ)
\[ V = 10\,e^{-(5.0/2.0)} = 10\,e^{-2.5} \approx 0.82\,\mathrm{V} \]

After two and a half time constants the capacitor retains under a tenth of its starting voltage — the exponential decay is rapid once a couple of time constants pass.

Review

Chapter Summary

✓ EMF

\(\mathcal{E} = \dfrac{dW}{dq}\), in volts; the work per unit charge an emf device supplies (not a force).

✓ Single loop

\(i = \dfrac{\mathcal{E}}{R + r}\); terminal voltage \(V = \mathcal{E} - ir\).

✓ Series

\(R_{\text{eq}} = \sum R_{j}\); same current through each, voltages add.

✓ Parallel

\(\dfrac{1}{R_{\text{eq}}} = \sum \dfrac{1}{R_{j}}\); same voltage across each, currents add.

✓ Kirchhoff's rules

Junction: \(\sum i_{\text{in}} = \sum i_{\text{out}}\) (charge). Loop: \(\sum V = 0\) (energy).

✓ Meters

Ammeter in series, low \(R\); voltmeter in parallel, high \(R\).

✓ Power

Supplied \(P = i\mathcal{E}\); lost internally \(P = i^{2}r\); to load \(P = i^{2}R\).

✓ RC circuit

Charge: \(q = C\mathcal{E}(1 - e^{-t/RC})\); discharge: \(q = q_{0}e^{-t/RC}\); \(\tau = RC\).

Practice

Problems

Treat batteries as real (with internal resistance \(r\)) only where one is stated. Keep straight which elements share a current (series) and which share a voltage (parallel), and choose loop-walking directions before writing any Kirchhoff equation.

  1. A 9.0 V battery of internal resistance 0.30 Ω is connected across a 4.5 Ω resistor. Find the current and the terminal voltage.
  2. What internal resistance would cause a 12 V battery to deliver a terminal voltage of 11.4 V while supplying 3.0 A?
  3. Four 100 Ω resistors are wired (a) all in series and (b) all in parallel. Find the equivalent resistance in each case.
  4. A 6.0 Ω and a 3.0 Ω resistor are in parallel, and that combination is in series with a 4.0 Ω resistor. Find the total resistance, and the current drawn from an ideal 10 V battery.
  5. In the previous problem, find the voltage across the parallel combination and the current through each of its two branches.
  6. Two batteries, 12 V and 6.0 V, with negligible internal resistance, are connected with their positive terminals together through a 2.0 Ω resistor. Find the current and its direction.
  7. A 3.0 A current enters a junction and splits into two branches of 6.0 Ω and 12 Ω in parallel. Find the current in each branch.
  8. Apply Kirchhoff's rules to a two-loop circuit: a 10 V battery in the left branch, a 5.0 V battery in the right branch (opposing), and a shared central 4.0 Ω resistor, with 2.0 Ω in each battery branch. Set up the equations for the three branch currents.
  9. An ideal voltmeter reads 6.0 V across a resistor carrying 0.50 A. Find the resistance and the power it dissipates.
  10. An ammeter of resistance 0.10 Ω is placed in series in a branch that nominally carries 2.0 A through a 5.0 Ω resistor on a 10 V supply. By what percentage does the meter reduce the current?
  11. A 5.0 µF capacitor charges through a 1.0 MΩ resistor from a 20 V source. Find the time constant, the final charge, and the current at \(t = 0\).
  12. For the circuit of the previous problem, how long until the capacitor reaches 90% of its final charge?
  13. A 10 µF capacitor charged to 50 V discharges through a 25 kΩ resistor. Find the time constant and the charge remaining after 1.0 s.
  14. Derive the discharging law \(q = q_{0}e^{-t/RC}\) from the loop rule \(q/C + R\,dq/dt = 0\), and show that \(\tau = RC\) has units of seconds.
Tip: three habits keep circuit problems clean. First, before touching algebra, reduce every series and parallel block you can to shrink the network — then restore the detail only where the question asks. Second, when you reach for Kirchhoff, commit to assumed current directions and a loop-walking direction first, then apply the sign rules (drop \(iR\) going with the current, rise \(\mathcal{E}\) from \(-\) to \(+\)) mechanically; a negative answer just means a guessed direction was backward. Third, for any RC question, find \(\tau = RC\) first and read off the limits — at \(t = 0\) an uncharged capacitor is a wire and a charged one a battery; at \(t \to \infty\) a capacitor is an open break.