Resistor-Capacitor (RC) Circuits: Study Guide (Physics 2 11.8)

Capacitors in combination

In Topic 10.6 a capacitor stored charge +Q on one plate and −Q on the other, with Q = CΔV. In a circuit, several capacitors can be replaced by one equivalent capacitance C_eq that stores the same charge for the same potential difference.

Parallel. Each capacitor is connected across the same two points, so each has the same ΔV. The charges add: Q = C₁ΔV + C₂ΔV. So:

C_eq = C₁ + C₂ + …

Adding a capacitor in parallel is like making the plates bigger, so C_eq goes up.

Series. Look at the inner plates of two capacitors in series: the right plate of C₁ and the left plate of C₂, joined by a wire (the dashed box in Figure 1). Nothing else connects to them, and they start uncharged. Charge is conserved, so their total charge stays zero. If −Q moves onto one, +Q must be left on the other. So every capacitor in series carries the same charge Q. The potential differences add: ΔV = Q/C₁ + Q/C₂. So:

1/C_eq = 1/C₁ + 1/C₂ + …

Because each term 1/C is positive, 1/C_eq is larger than any single 1/C. So C_eq is smaller than the smallest capacitor in the series. For example, two 2.0 μF capacitors give 1.0 μF in series and 4.0 μF in parallel.

Notice the pattern is the reverse of resistors: series resistances add, but parallel capacitances add.

Charging a capacitor through a resistor

Connect an uncharged capacitor C, a resistor R, a switch and an ideal battery of emf ℰ in one loop, then close the switch. The loop rule holds at every instant:

ℰ = ΔV_R + ΔV_C = IR + Q/C

Follow what happens.

• Just after closing. Q = 0, so ΔV_C = 0. The capacitor acts like a plain wire: charge flows easily onto its plates. The whole emf is across the resistor, so the current is largest: I₀ = ℰ/R.
• While charging. Charge builds up, so ΔV_C = Q/C rises. Then ΔV_R = ℰ − ΔV_C falls, so the current falls. The stored energy U = ½QΔV_C rises.
• After a long time. ΔV_C approaches ℰ. Then ΔV_R approaches 0 and the current approaches zero. The capacitor holds its maximum charge Q = Cℰ and acts like a break in its branch.

Each quantity changes quickly at first, then more and more slowly, approaching its final value asymptotically.

The current is in the wires on both sides of the capacitor: charge flows onto one plate and the same amount flows off the other. No charge crosses the gap.

Discharging a capacitor

Now connect a charged capacitor (potential difference ΔV₀) across a resistor. The capacitor drives the current, so the current in its branch flows the opposite way to the charging current.

• Just after the connection, ΔV_R = ΔV₀, so the current is largest, ΔV₀/R. At once, the charge and stored energy start to fall.
• While discharging, Q, ΔV_C and I all decrease together, quickly at first and then more slowly.
• After a long time, all three approach zero.

The time constant τ = RC

How fast these changes happen depends on the time constant:

τ = RC

Units: Ω × F = (V/A) × (C/V) = C/A = s. So τ is a time.

• Charging: after one time constant, the charge (and ΔV_C) has reached about 63% of its final value.
• Discharging: after one time constant, the charge (and ΔV_C and I) has fallen to about 37% of its starting value.

Why RC? A larger R means a smaller current for the same potential difference, so charge moves more slowly. A larger C means more charge must move to reach the same potential difference. Either way the process takes longer. The emf does not change τ: a bigger emf means more charge to move, but also a proportionally bigger current.

Background, not assessed: the 63% and 37% figures come from an exponential model (37% ≈ 1/e). After about 5τ the capacitor is within 1% of its final state, which is why “after a long time” means many time constants.

Capacitors in circuits with branches

For circuits with more than one branch, use two snapshots and Kirchhoff’s rules.

1. Just after a switch closes: replace each uncharged capacitor with a wire. Solve for the currents.
2. After a long time: replace each capacitor with a break. Its branch carries no current. Solve for the currents, then find ΔV_C from the loop rule. ΔV_C equals the potential difference across whatever the capacitor’s branch is connected in parallel with.

Between these two states, every quantity changes smoothly from its first value to its last.

Worked example 1: equivalent capacitance

Question. C₁ = 6.0 μF is connected in series with a parallel pair, C₂ = 1.0 μF and C₃ = 2.0 μF. The combination is connected to a 9.0 V battery and fully charged. Find C_eq, the charge and potential difference for each capacitor, and the total stored energy.

1. Parallel pair: C₂₃ = 1.0 + 2.0 = 3.0 μF.
2. In series with C₁: 1/C_eq = 1/6.0 + 1/3.0 = 1/2.0, so C_eq = 2.0 μF. (Smaller than 3.0 μF, as it must be.)
3. Total charge: Q = C_eq ΔV = (2.0 μF)(9.0 V) = 18 μC. C₁ and the pair are in series, so each carries 18 μC.
4. ΔV₁ = 18 μC ÷ 6.0 μF = 3.0 V. ΔV₂₃ = 18 μC ÷ 3.0 μF = 6.0 V.
5. Q₂ = (1.0 μF)(6.0 V) = 6.0 μC; Q₃ = (2.0 μF)(6.0 V) = 12 μC.
6. Energy: U = ½C_eq(ΔV)² = ½(2.0 × 10⁻⁶ F)(9.0 V)² = 81 μJ.

Check. 3.0 V + 6.0 V = 9.0 V (loop rule). 6.0 μC + 12 μC = 18 μC (charge conservation). Energy by parts: ½(18 μC)(3.0 V) + ½(18 μC)(6.0 V) = 27 + 54 = 81 μJ.

Worked example 2: start and end states

Question. An ideal 12 V battery, a switch and R₁ = 2.0 kΩ are in series. After R₁ the circuit splits into two parallel branches: an uncharged capacitor C = 50 μF, and a resistor R₂ = 4.0 kΩ. The switch is closed at t = 0. Find the currents just after closing and after a long time, and the final charge and energy on C.

1. Just after closing, C acts like a wire. It is in parallel with R₂, so R₂ is short-circuited: ΔV across R₂ is 0 and I_R₂ = 0.
2. All the emf is across R₁: I₁ = 12 V ÷ 2000 Ω = 6.0 mA. By the junction rule, I_C = I₁ − I_R₂ = 6.0 mA.
3. After a long time, C acts like a break, so I_C = 0. The junction rule gives I_R₂ = I₁.
4. One loop through R₁ and R₂: I₁ = 12 V ÷ 6000 Ω = 2.0 mA, so I_R₂ = 2.0 mA.
5. ΔV_C = ΔV across R₂ = (2.0 mA)(4.0 kΩ) = 8.0 V, not 12 V. The other 4.0 V is across R₁.
6. Q = CΔV_C = (50 μF)(8.0 V) = 400 μC. U = ½C(ΔV_C)² = ½(50 × 10⁻⁶ F)(8.0 V)² = 1.6 mJ.

Interpretation. The battery current falls from 6.0 mA to 2.0 mA while the R₂ current rises from 0 to 2.0 mA. The capacitor only reaches the potential difference of the element it is in parallel with.

Worked example 3: finding τ from discharge data

Question. A student charges a capacitor, then discharges it through a 60 kΩ resistor, recording ΔV_C (fictional data): 8.0 V at 0 s, 5.3 V at 5 s, 3.5 V at 10 s, 2.3 V at 15 s, 1.5 V at 20 s, 1.0 V at 25 s and 0.66 V at 30 s. Estimate τ and the capacitance.

1. Plot ΔV_C against t with even scales and draw a smooth curve (Figure 3).
2. After one time constant, ΔV_C is about 37% of its starting value: 0.37 × 8.0 V = 2.96 V.
3. 2.96 V lies between 3.5 V (10 s) and 2.3 V (15 s). Reading from the curve, or interpolating, gives τ ≈ 12 s.
4. C = τ/R = 12 s ÷ (60 × 10³ Ω) = 2.0 × 10⁻⁴ F = 200 μF.

Check. After another 12 s (t ≈ 24 s), ΔV_C should be about 37% of 2.96 V, roughly 1.1 V. The data give about 1.1 V between 20 s and 25 s. To improve the estimate, repeat the run and measure the time to fall to 37% from several starting points.

Common misconceptions

• “Capacitors combine like resistors.” The rules swap: parallel capacitances add; series capacitances combine as reciprocals.
• “Capacitors in series have different charges.” They all carry the same Q. It is their ΔV values that differ, with the largest ΔV across the smallest capacitor.
• “An uncharged capacitor blocks current at first.” The reverse: at first it acts like a wire. It blocks steady current only once it is charged.
• “Charge flows through the capacitor.” Charge flows onto one plate and off the other. None crosses the gap.
• “The capacitor is full after one time constant.” After τ it has about 63% of its final charge. It approaches full charge only after many time constants.
• “A bigger battery charges the capacitor faster.” τ = RC does not depend on the emf.
• “The final ΔV_C always equals the battery emf.” Only if no current flows in the rest of the loop. In Worked example 2 it is 8.0 V, not 12 V.
• “The current in a discharging capacitor flows the same way as when charging.” The capacitor now drives the current, so in its branch it flows the other way.

Where this leads

This completes the circuits unit. Next, Topic 12.1, Magnetic Fields, starts magnetism, where moving charges and currents produce and feel magnetic forces. First test yourself with the practice questions, then use the revision notes and the topic checklist to consolidate. To revise the junction rule used in Worked example 2, return to Topic 11.7.