Kirchhoff's rules, RC circuits, ammeter and voltmeter

The last topic of the semester and the heaviest part of the final exam. The circuits you solved last week by series and parallel reduction only go so far: a circuit with two batteries in different branches cannot be reduced. For this there are Kirchhoff’s two rules. After that a capacitor enters the circuit and the current now changes with time. Expect three question types on the exam: a branch current in a two-loop circuit, the charge or voltage at a given moment in an RC circuit, and the design of an ammeter/voltmeter.

Golden rules:

1. At a junction, current in = current out. Charge does not accumulate.
2. Around a closed path the potential changes sum to zero. When you return to the point you started from, you are at the same potential.
3. Choose the current directions arbitrarily at the start. If a result comes out negative, the real direction is opposite; your equations are still right.
4. $\tau = RC$. After one $\tau$, a charging capacitor has reached 63% of its final charge and a discharging one has 37% of its initial charge left.
5. An ammeter is in series and has a small resistance, a voltmeter is in parallel and has a large resistance.

Kirchhoff’s rules

Junction rule. At any junction (a point where three or more wires meet):

$\sum I_{\text{in}} = \sum I_{\text{out}}$

Loop rule. Along any closed path:

$\sum_{\text{loop}} \Delta V = 0$

When applying the loop rule, decide whether the potential rises or falls at each element using this table:

PITFALL · the sign of a battery does not depend on the current. For a battery, the only thing that sets the sign is your direction of traversal: if you go from negative to positive it is $+\varepsilon$. The direction the current passes through the battery is irrelevant. For a resistor, the current direction sets the sign.

Method

1. Assign a current ($I_1, I_2, \ldots$) and a direction to each branch. Series elements are in the same branch and carry the same current.
2. Write the junction rule (number of junctions − 1) times.
3. Write the remaining equations from independent loops (usually as many as the number of “windows” in the circuit).
4. Solve the system of equations. A current that comes out negative actually flows the other way.
5. Check: the sum of the powers supplied by the batteries = the power dissipated in the resistors.

Worked example: a circuit with two batteries

The circuit: there is a node a at the top and a node b at the bottom, with three parallel branches between them.

• Left branch: $\varepsilon_1 = 12$ V (positive terminal facing a) and $R_1 = 2.0\ \Omega$ in series.
• Middle branch: only $R_2 = 4.0\ \Omega$.
• Right branch: $\varepsilon_2 = 6.0$ V (positive terminal facing a) and $R_3 = 3.0\ \Omega$ in series.

Currents: $I_1$ in the left branch from b to a, $I_3$ in the right branch from b to a, $I_2$ in the middle branch from a to b.

Node a: $I_1 + I_3 = I_2$

Left loop (start at b, go up the left branch, down the middle branch): in the battery from negative to positive $+12$, in $R_1$ along the current $-2.0I_1$, in $R_2$ along the current $-4.0I_2$:

$12 - 2.0I_1 - 4.0I_2 = 0$

Right loop (start at b, go up the right branch, down the middle branch):

$6.0 - 3.0I_3 - 4.0I_2 = 0$

Solution: from the first two loop equations $I_1 = 6.0 - 2.0I_2$ and $I_3 = 2.0 - \frac{4}{3}I_2$. Put into the node equation:

6.0 − 2.0 I2 + 2.0 − 1.333 I2 = I2
8.0 = 4.333 I2   →   I2 = 1.85 A
I1 = 6.0 − 2(1.85) = 2.31 A
I3 = 2.0 − 1.333(1.85) = −0.46 A

$I_3$ came out negative: in the right branch the current actually flows from a to b, that is, into the positive terminal of the 6.0 V battery. The big battery is charging the small one.

Power check:

Battery 1 supplies:    12 × 2.31  = 27.7 W
Battery 2 absorbs:     6.0 × 0.46 =  2.8 W   (being charged)
Net:                   24.9 W
Resistors:             2(2.31)² + 4(1.85)² + 3(0.46)² = 10.7 + 13.6 + 0.6 = 24.9 W

TIP · node potential shortcut. If there are only two nodes, take b as 0 V, call the potential of a $V$, and write a single equation: the current in each branch is $(\varepsilon - V)/R$. In this example $\frac{12 - V}{2} + \frac{6 - V}{3} = \frac{V}{4}$ immediately gives $V = 7.38$ V. One equation instead of a system of three unknowns.

RC circuits

In a circuit where a resistor and a capacitor are in series, the current is not constant. As the capacitor charges, the voltage $q/C$ across it grows, the voltage left for the resistor shrinks, and the current falls.

Charging a capacitor

An initially empty capacitor, a battery with emf $\varepsilon$ and a resistor $R$ in series. The switch closes at $t = 0$. Loop rule:

$\varepsilon - iR - \frac{q}{C} = 0, \qquad i = \frac{dq}{dt}$

The solution of this differential equation is:

$q(t) = C\varepsilon\left(1 - e^{-t/RC}\right), \qquad i(t) = \frac{\varepsilon}{R}e^{-t/RC}$

$\tau = RC$ is the time constant, in seconds. The important points:

Example. $R = 50$ kΩ, $C = 20\ \mu$F, $\varepsilon = 10$ V.

τ = RC = (50e3)(20e-6) = 1.0 s
i(0) = ε/R = 10 / 50e3 = 0.20 mA
Q_final = Cε = 20 μF × 10 V = 200 μC
t = 2.0 s:  ΔV_C = 10 (1 − e^−2) = 8.6 V
90% charged:  e^(−t/τ) = 0.10  →  t = τ ln 10 = 2.3 s

Discharging a capacitor

A capacitor with charge $Q$ discharges through $R$ with no battery:

$q(t) = Q\,e^{-t/RC}, \qquad i(t) = -\frac{Q}{RC}e^{-t/RC}$

The minus sign says that the current reduces the charge of the capacitor. The time for the charge to halve is $t_{1/2} = \tau\ln 2 = 0.693\,\tau$.

PITFALL · energy does not fall at the same rate as charge. Since $U = q^2/2C$, $U \propto e^{-2t/\tau}$. It takes $\tau\ln2/2$ for the energy to halve, half the time it takes for the charge to halve. On the exam both options are present.

Steady-state questions

If a circuit with a capacitor has waited a very long time (steady state), no current flows in the capacitor branch. First “disconnect” the capacitor from the circuit, solve the remaining resistor circuit, find the voltage between the two points the capacitor is connected to, and compute $Q = C\Delta V$. If the switch has just closed ($t = 0^+$), think of an empty capacitor as a wire.

Example. An ideal 12 V battery, $R_1 = 4.0\ \Omega$ and $R_2 = 2.0\ \Omega$ in series; $C = 3.0\ \mu$F connected directly across $R_2$.

t = 0:   C is a short circuit, bridging R2   → I = 12/4.0 = 3.0 A
t → ∞:   no current in the C branch          → I = 12/6.0 = 2.0 A
         ΔV_C = ΔV_R2 = 2.0 × 2.0 = 4.0 V
         Q = 3.0 μF × 4.0 V = 12 μC

Galvanometer, ammeter, voltmeter

A galvanometer is a sensitive instrument whose needle deflects with a small current through it. It is defined by two things: the coil resistance $R_g$ and the full-scale deflection current $I_g$ (typically of the order of mA).

An ammeter measures current and is connected in series with the branch it measures. Its resistance must be very small so that it does not disturb the circuit. A small shunt resistor is connected in parallel with the galvanometer; the excess current passes through the shunt:

$I_gR_g = (I - I_g)R_{sh} \quad\Rightarrow\quad R_{sh} = \frac{I_gR_g}{I - I_g}$

A voltmeter measures a potential difference and is connected in parallel with the element it measures. Its resistance must be very large so that it does not steal current from the circuit. A large resistor is added in series with the galvanometer:

$V = I_g(R_g + R_s) \quad\Rightarrow\quad R_s = \frac{V}{I_g} - R_g$

Loading effect. Real instruments are not ideal. If you connect a 0.50 Ω ammeter in series in a circuit with a 6.0 V battery and a 3.0 Ω resistor, you read $6.0/3.5 = 1.7$ A instead of 2.0 A. Similarly, if you connect a 10 kΩ voltmeter across one of two 10 kΩ resistors, that part drops to 5.0 kΩ and you read 4.0 V instead of 6.0 V. This effect grows when the resistance of the instrument is comparable to the resistances in the circuit you measure.

Common mistakes