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Capacitor question

2022 · 26 Jul · Shift 1 · Q54
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  5. /2022 · 26 Jul · Shift 1 · Q54

Capacitor question

2022 · 26 Jul · Shift 1 · Q54

JEE MainPhysicsCapacitorMCQ+4 / −1
The total charge on the system of capacitors C1=1μF,C2=2μF,C3=4μFC_{1}=1 \mu \mathrm{F}, C_{2}=2 \mu \mathrm{F}, \mathrm{C}_{3}=4 \mu \mathrm{F}C1​=1μF,C2​=2μF,C3​=4μF and C4=3μF\mathrm{C}_{4}=3 \mu \mathrm{F}C4​=3μF connected in parallel is : (Assume a battery of 20 V20 \mathrm{~V}20 V is connected to the combination)
  1. A
    200 μC200 \,\mu \mathrm{C}200μC
  2. B
    200 C
  3. C
    10 μC10 \,\mu \mathrm{C}10μC
  4. D
    10 C
View written solutionFree

Correct answer: A

  1. Capacitors in parallel

    For capacitors connected in parallel, the equivalent capacitance is the sum: Ceq=C1+C2+C3+C4C_{\text{eq}}=C_1+C_2+C_3+C_4Ceq​=C1​+C2​+C3​+C4​

    Given: C1=1 μF,C2=2 μF,C3=4 μF,C4=3 μFC_1=1\,\mu F,\quad C_2=2\,\mu F,\quad C_3=4\,\mu F,\quad C_4=3\,\mu FC1​=1μF,C2​=2μF,C3​=4μF,C4​=3μF

    Therefore, Ceq=1+2+4+3=10 μFC_{\text{eq}}=1+2+4+3=10\,\mu FCeq​=1+2+4+3=10μF

  2. Use the relation for charge

    Total charge on the combination is: Q=CeqVQ=C_{\text{eq}}VQ=Ceq​V

    Given battery voltage: V=20 VV=20\,VV=20V

    So, Q=(10 μF)(20 V)=200 μCQ=(10\,\mu F)(20\,V)=200\,\mu CQ=(10μF)(20V)=200μC

  3. Check options

    • A: 200 μC200\,\mu C200μC ✅
    • B: 200 C200\,C200C ❌
    • C: 10 μC10\,\mu C10μC ❌
    • D: 10 C10\,C10C ❌

Thus, the correct answer is A.

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