Capacitors: Speed-Solving Techniques (3)

hard • 3 min read
Tags Capacitors

Question

Three capacitors C1=2 μC_1 = 2\,\muF, C2=3 μC_2 = 3\,\muF, and C3=6 μC_3 = 6\,\muF are connected. C1C_1 and C2C_2 are in series; this combination is in parallel with C3C_3. The combination is connected across a 12 V battery. Find the equivalent capacitance, the total charge stored, and the voltage across C1C_1.

Solution — Step by Step

For two capacitors in series, C12=C1C2C1+C2=2×32+3=1.2 μC_{12} = \dfrac{C_1 C_2}{C_1 + C_2} = \dfrac{2 \times 3}{2 + 3} = 1.2\,\muF. This is the product-over-sum trick — works only for two capacitors in series.

In parallel, capacitances add directly: Ceq=C12+C3=1.2+6=7.2 μC_{\text{eq}} = C_{12} + C_3 = 1.2 + 6 = 7.2\,\muF.

Qtotal=CeqV=7.2 μF×12 V=86.4 μCQ_{\text{total}} = C_{\text{eq}} V = 7.2\,\mu\text{F} \times 12 \text{ V} = 86.4\,\mu\text{C}

C1C_1 and C2C_2 are in series, so they carry the same charge. Voltage across the series combination = 12 V. Charge on C12=1.2×12=14.4 μC_{12} = 1.2 \times 12 = 14.4\,\muC. So voltage across C1C_1:

V1=QC1=14.42=7.2 VV_1 = \frac{Q}{C_1} = \frac{14.4}{2} = 7.2 \text{ V}

Why This Works

Capacitors in series share the same charge but split the voltage; capacitors in parallel share the same voltage but split the charge. These two facts unlock every capacitor network problem. Don’t memorize the series formula by rote — derive it: V1+V2=VV_1 + V_2 = V, Q/C1+Q/C2=VQ/C_1 + Q/C_2 = V, so Q=V⋅C1C2C1+C2Q = V \cdot \dfrac{C_1 C_2}{C_1 + C_2}.

For series combinations of two, the voltage across C1C_1 is V⋅C2C1+C2V \cdot \dfrac{C_2}{C_1 + C_2}. Quick check: 12×3/5=7.212 \times 3/5 = 7.2 V. ✓

Three speed shortcuts for capacitor circuits:

  1. Two in series: Ceq=C1C2/(C1+C2)C_{\text{eq}} = C_1 C_2 / (C_1 + C_2). Voltage across C1C_1: V⋅C2/(C1+C2)V \cdot C_2/(C_1+C_2) — note the cross-multiplication.
  2. Two in parallel: Ceq=C1+C2C_{\text{eq}} = C_1 + C_2. Charge on C1C_1: V⋅C1V \cdot C_1.
  3. Energy stored: U=12CV2=Q22C=12QVU = \tfrac{1}{2} C V^2 = \tfrac{Q^2}{2C} = \tfrac{1}{2} Q V — pick the form using the variables you have.

Alternative Method

Voltage divider for series capacitors: V1=V⋅C2C1+C2=12⋅35=7.2V_1 = V \cdot \dfrac{C_2}{C_1 + C_2} = 12 \cdot \dfrac{3}{5} = 7.2 V. Note this is the opposite of the resistor voltage divider — bigger capacitance means smaller voltage share, because V=Q/CV = Q/C.

Students apply the resistor voltage divider V1=VC1/(C1+C2)V_1 = V C_1/(C_1 + C_2) to capacitors. Wrong. For capacitors in series, VV is split inversely with CC: bigger capacitor takes less voltage.

Final answer: Ceq=7.2 μC_{\text{eq}} = 7.2\,\muF, Q=86.4 μQ = 86.4\,\muC, V1=7.2V_1 = 7.2 V.

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