Question

Difficulty: MediumCapacitors and Capacitance

Two capacitors of capacitances 8.0 μF8.0\text{ }\mu\text{F} and 4.0 μF4.0\text{ }\mu\text{F} are connected in parallel. This parallel combination is then connected in series with a single 4.0 μF4.0\text{ }\mu\text{F} capacitor across a 36.0 V36.0\text{ V} d.c. voltage source. What is the potential difference across the single 4.0 μF4.0\text{ }\mu\text{F} capacitor?

  1. A
    9.0 V9.0\text{ V}
  2. B
    14.4 V14.4\text{ V}
  3. 27.0 V27.0\text{ V}Answer
  4. D
    36.0 V36.0\text{ V}

Answer

The potential difference across the single 4.0 μF4.0\text{ }\mu\text{F} capacitor is 27.0 V27.0\text{ V}.
The two parallel capacitors (8.0 μF8.0\text{ }\mu\text{F} and 4.0 μF4.0\text{ }\mu\text{F}) combine directly to give an equivalent capacitance of 12.0 μF12.0\text{ }\mu\text{F}. This equivalent capacitor is in series with the single 4.0 μF4.0\text{ }\mu\text{F} capacitor across the 36.0 V36.0\text{ V} source. Using the voltage divider rule for series capacitors, the voltage across the single 4.0 μF4.0\text{ }\mu\text{F} capacitor is V=36.0×12.04.0+12.0=27.0 VV = 36.0 \times \frac{12.0}{4.0 + 12.0} = 27.0\text{ V}.

Step-by-Step Solution

1
Calculate the equivalent capacitance of the two parallel capacitors.
Cp=8.0 μF+4.0 μF=12.0 μFC_p = 8.0\text{ }\mu\text{F} + 4.0\text{ }\mu\text{F} = 12.0\text{ }\mu\text{F}
Capacitors in parallel add algebraically.
2
Calculate the total equivalent capacitance of the entire circuit.
CT=4.0×12.04.0+12.0=48.016.0=3.0 μFC_T = \frac{4.0 \times 12.0}{4.0 + 12.0} = \frac{48.0}{16.0} = 3.0\text{ }\mu\text{F}
The single 4.0 μF4.0\text{ }\mu\text{F} capacitor and the 12.0 μF12.0\text{ }\mu\text{F} parallel combination are in series.
3
Find the total charge supplied by the battery.
Q=CT×V=3.0 μF×36.0 V=108.0 μCQ = C_T \times V = 3.0\text{ }\mu\text{F} \times 36.0\text{ V} = 108.0\text{ }\mu\text{C}
The total charge is equal to the total capacitance multiplied by the total voltage.
4
Calculate the potential difference across the single 4.0 μF4.0\text{ }\mu\text{F} capacitor.
V1=QC1=108.0 μC4.0 μF=27.0 VV_1 = \frac{Q}{C_1} = \frac{108.0\text{ }\mu\text{C}}{4.0\text{ }\mu\text{F}} = 27.0\text{ V}
In a series connection, the charge on the single capacitor equals the total charge.

Key Concept

Potential difference distribution in mixed series-parallel capacitor networks
Estimated Time:1m 30s
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