Question:

A capacitor, C1=6μF C_1 = 6 \, \mu F , is charged to a potential difference of V1=5V V_1 = 5 \, \text{V} using a 5V battery. The battery is removed and another capacitor, C2=12μF C_2 = 12 \, \mu F , is inserted in place of the battery. When the switch 'S' is closed, the charge flows between the capacitors for some time until equilibrium condition is reached. What are the charges q1 q_1 and q2 q_2 on the capacitors C1 C_1 and C2 C_2 when equilibrium condition is reached?
battery is removed and another capacitor

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When capacitors are in parallel and the switch is closed, the total charge is conserved, and the potential across all capacitors will be the same at equilibrium. Use charge conservation and the capacitance values to find the final charges on each capacitor.
Updated On: Mar 18, 2025
  • q1=15μC,q2=30μC q_1 = 15 \, \mu C, \, q_2 = 30 \, \mu C
  • q1=30μC,q2=15μC q_1 = 30 \, \mu C, \, q_2 = 15 \, \mu C
  • q1=10μC,q2=20μC q_1 = 10 \, \mu C, \, q_2 = 20 \, \mu C
  • q1=20μC,q2=10μC q_1 = 20 \, \mu C, \, q_2 = 10 \, \mu C
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The Correct Option is C

Solution and Explanation

Step 1: At t=0 t = 0 , the initial charge on C1 C_1 is: q1=C1V1=6μF5V=30μC q_1 = C_1 \cdot V_1 = 6 \, \mu F \cdot 5 \, \text{V} = 30 \, \mu C

Step 2: After the switch 'S' is closed, charge flows until equilibrium is reached, and the total charge is distributed between the two capacitors. The final charge on each capacitor can be found using the conservation of charge and voltage. At equilibrium, the potential difference across both capacitors will be the same. Let Vc V_c be the common potential difference at equilibrium. q1=C1Vcandq2=C2Vc q_1 = C_1 \cdot V_c \quad \text{and} \quad q_2 = C_2 \cdot V_c Using the total charge conservation: q1+q2=30μC(total charge is conserved) q_1 + q_2 = 30 \, \mu C \quad \text{(total charge is conserved)} Substitute the expressions for q1 q_1 and q2 q_2 : C1Vc+C2Vc=30μC C_1 \cdot V_c + C_2 \cdot V_c = 30 \, \mu C Vc(C1+C2)=30μC V_c \cdot (C_1 + C_2) = 30 \, \mu C Now, solve for Vc V_c : Vc=30μCC1+C2=30μC6μF+12μF=30μC18μF=1.67V V_c = \frac{30 \, \mu C}{C_1 + C_2} = \frac{30 \, \mu C}{6 \, \mu F + 12 \, \mu F} = \frac{30 \, \mu C}{18 \, \mu F} = 1.67 \, \text{V}

Step 3: Finally, the charges on the capacitors are: q1=C1Vc=6μF1.67V=10μC q_1 = C_1 \cdot V_c = 6 \, \mu F \cdot 1.67 \, \text{V} = 10 \, \mu C q2=C2Vc=12μF1.67V=20μC q_2 = C_2 \cdot V_c = 12 \, \mu F \cdot 1.67 \, \text{V} = 20 \, \mu C Thus, the charges are q1=10μC q_1 = 10 \, \mu C and q2=20μC q_2 = 20 \, \mu C , so the correct answer is option (3).
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