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Electrostatic Potential and Capacitance - Combination of Capacitors

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Capacitors are in Series when they are connected end-to-end such that the same charge QQ flows through each capacitor. The total potential difference VV is the sum of individual potential differences: V=V1+V2+V3V = V_1 + V_2 + V_3.

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In a series combination, the equivalent capacitance CsC_s is always smaller than the smallest individual capacitance in the network.

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Capacitors are in Parallel when they are connected across the same two points, meaning the potential difference VV is the same across each capacitor. The total charge QQ is the sum of individual charges: Q=Q1+Q2+Q3Q = Q_1 + Q_2 + Q_3.

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In a parallel combination, the equivalent capacitance CpC_p is the sum of individual capacitances and is larger than the largest individual capacitance.

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Energy stored in a capacitor is given by U=12CV2U = \frac{1}{2}CV^2. When capacitors are combined, the total energy stored is the sum of the energies stored in individual capacitors, regardless of series or parallel connection.

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For nn identical capacitors each of capacitance CC: In series, Cs=CnC_s = \frac{C}{n}; In parallel, Cp=nCC_p = nC.

📐Formulae

1Cs=1C1+1C2+1C3+⋯+1Cn\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \dots + \frac{1}{C_n}

Cp=C1+C2+C3+⋯+CnC_p = C_1 + C_2 + C_3 + \dots + C_n

Q=CVQ = CV

V=V1+V2+⋯+Vn (Series)V = V_1 + V_2 + \dots + V_n \text{ (Series)}

Q=Q1+Q2+⋯+Qn (Parallel)Q = Q_1 + Q_2 + \dots + Q_n \text{ (Parallel)}

U=12CV2=Q22C=12QVU = \frac{1}{2}CV^2 = \frac{Q^2}{2C} = \frac{1}{2}QV

💡Examples

Problem 1:

Three capacitors of capacitances 2μF2\mu F, 3μF3\mu F, and 6μF6\mu F are connected in series to a 12V12V battery. Calculate the equivalent capacitance and the charge on each capacitor.

Solution:

  1. Equivalent capacitance CsC_s: 1Cs=12+13+16\frac{1}{C_s} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} 1Cs=3+2+16=66=1μF\frac{1}{C_s} = \frac{3 + 2 + 1}{6} = \frac{6}{6} = 1\mu F So, Cs=1μFC_s = 1\mu F.
  2. Total charge QQ: Q=Cs×V=1μF×12V=12μCQ = C_s \times V = 1\mu F \times 12V = 12\mu C. Since they are in series, the charge on each capacitor is the same: Q1=Q2=Q3=12μCQ_1 = Q_2 = Q_3 = 12\mu C.

Explanation:

In a series circuit, the reciprocal of the total capacitance is the sum of the reciprocals of individual capacitances. The charge remains constant across all capacitors in series.

Problem 2:

Two capacitors of 4μF4\mu F and 12μF12\mu F are connected in parallel. If the total charge supplied to the combination is 160μC160\mu C, find the potential difference across the combination and the charge on each capacitor.

Solution:

  1. Equivalent capacitance CpC_p: Cp=4μF+12μF=16μFC_p = 4\mu F + 12\mu F = 16\mu F
  2. Potential difference VV: V=QtotalCp=160μC16μF=10VV = \frac{Q_{total}}{C_p} = \frac{160\mu C}{16\mu F} = 10V
  3. Charge on each: Q1=C1V=4μF×10V=40μCQ_1 = C_1V = 4\mu F \times 10V = 40\mu C Q2=C2V=12μF×10V=120μCQ_2 = C_2V = 12\mu F \times 10V = 120\mu C Verification: 40+120160\begin{array}{r} 40 \\ + 120 \\ \hline 160 \end{array}

Explanation:

In parallel, capacitances add up directly. The potential difference is the same for both, and the total charge is distributed based on the capacitance values (Q=CVQ = CV).