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Concepts Of Physics MCQ Edition [Volume 2]
Exam
Concepts Of Physics MCQ Edition [Volume 2]
Subject
Physics
Chapter
Capacitors

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Capacitors

Three capacitors with capacitances of 20 , 30 , and 40 are...

Three capacitors with capacitances of 20μF20 \mu\text{F}, 30μF30 \mu\text{F}, and 40μF40 \mu\text{F} are connected in series across a 12 V12 \text{ V} battery. Determine the charge acquired by each capacitor and the work done by the battery during the charging process.

Options

  1. A1080μC1080 \mu\text{C} on each, 1.30×10−2 J1.30 \times 10^{-2} \text{ J}
  2. B144013μC\dfrac{1440}{13} \mu\text{C} on each, 1.33×10−3 J1.33 \times 10^{-3} \text{ J}
  3. C48013μC\dfrac{480}{13} \mu\text{C} on each, 1.33×10−3 J1.33 \times 10^{-3} \text{ J}
  4. D144013μC\dfrac{1440}{13} \mu\text{C} on each, 6.65×10−4 J6.65 \times 10^{-4} \text{ J}

Correct answer

B. 144013μC\dfrac{1440}{13} \mu\text{C} on each, 1.33×10−3 J1.33 \times 10^{-3} \text{ J}

Step-by-step solution

The equivalent capacitance CeqC_{eq} for capacitors connected in series is given by:

1Ceq=1C1+1C2+1C3\dfrac{1}{C_{eq}} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \dfrac{1}{C_3}

1Ceq=120+130+140=6+4+3120=13120μ\textF−1\dfrac{1}{C_{eq}} = \dfrac{1}{20} + \dfrac{1}{30} + \dfrac{1}{40} = \dfrac{6 + 4 + 3}{120} = \dfrac{13}{120} \mu\textF^{-1}

Ceq=12013μFC_{eq} = \dfrac{120}{13} \mu\text{F}

In a series circuit, the charge QQ on each capacitor is the same and is equal to the total charge supplied by the battery:

Q=CeqV=(12013μF)×12 V=144013μCQ = C_{eq}V = \left(\dfrac{120}{13} \mu\text{F}\right) \times 12 \text{ V} = \dfrac{1440}{13} \mu\text{C}

The work done by the battery during the charging process is given by W=QVW = QV:

W=(144013×10−6 C)×12 V=1728013×10−6 JW = \left(\dfrac{1440}{13} \times 10^{-6} \text{ C}\right) \times 12 \text{ V} = \dfrac{17280}{13} \times 10^{-6} \text{ J}

W≈1329.23×10−6 J=1.33×10−3 JW \approx 1329.23 \times 10^{-6} \text{ J} = 1.33 \times 10^{-3} \text{ J}
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