Q18. A network of three capacitors each 9 µF connected in series and the fourth capacitor of 6 µF is

Q: What is the correct answer for CUET UG Physics Q18?

The correct answer is d. From the circuit diagram, capacitors C₁, C₂, and C₃ (each 9 µF) are connected in series. Their equivalent capacitance (C_series) is given by:1/C_series = 1/C₁ + 1/C₂ + 1/C₃ = 1/9µF + 1/9µF + 1/9µF = 3/9µF = 1/3µF.So, C_series = 3 µF.Capacitor C₄ (6 µF) is connected in parallel with this series combination. The entire parallel combination is supplied with 400 V.In a series combination, the charge a

Q: Q18. A network of three capacitors each 9 µF connected in series and the fourth capacitor of 6 µF is supplied 400 V as shown in figure. The ratio of charge on the capacitor C₁ to the capacitor C4…

The correct answer is d. From the circuit diagram, capacitors C₁, C₂, and C₃ (each 9 µF) are connected in series. Their equivalent capacitance (C_series) is given by:1/C_series = 1/C₁ + 1/C₂ + 1/C₃ = 1/9µF + 1/9µF + 1/9µF = 3/9µF = 1/3µF.So, C_series = 3 µF.Capacitor C₄ (6 µF) is connected in parallel with this series combination. The entire parallel combination is supplied with 400 V.In a series combination, the charge a

Q: What are the options for this CUET UG Physics question?

A. 1:3 · B. 1:4 · C. 4:1 · D. 1:2

Question

Q18. A network of three capacitors each 9 µF connected in series and the fourth capacitor of 6 µF is supplied 400 V as shown in figure. The ratio of charge on the capacitor C₁ to the capacitor C4 is:
(a) 1:3
(b) 1:4
(c) 4:1
(d) 1:2

FirstInTest · Accepted Answer

From the circuit diagram, capacitors C₁, C₂, and C₃ (each 9 µF) are connected in series. Their equivalent capacitance (C_series) is given by:1/C_series = 1/C₁ + 1/C₂ + 1/C₃ = 1/9µF + 1/9µF + 1/9µF = 3/9µF = 1/3µF.So, C_series = 3 µF.Capacitor C₄ (6 µF) is connected in parallel with this series combination. The entire parallel combination is supplied with 400 V.In a series combination, the charge across each capacitor is the same. Therefore, the charge on C₁ (Q₁) is equal to the total charge on the series combination (Q_series).Q₁ = Q_series = C_series × V = 3 µF × 400 V = 1200 µC.The voltage across C₄ (V₄) is the same as the supply voltage (V) because it's in parallel: V₄ = 400 V.Charge on C₄ (Q₄) = C₄ × V₄ = 6 µF × 400 V = 2400 µC.The ratio of charge on C₁ to C₄ is Q₁ : Q₄ = 1200 µC : 2400 µC = 1 : 2.

Answer

1:3

Answer

1:4

Answer

4:1

Answer

1:2

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