A 5pF capacitor is connected in series with the 10 PF capacitor and the combination across is 9V battery .the potential differences across the capacitor are in ratio . please answer with solution?

Question

Answer

Explanation:

Capacitors in Series:

When capacitors are connected in series, the equivalent capacitance is given by:

1/Ceq = 1/C1 + 1/C2 + ...

In this case, we have two capacitors in series:

1/Ceq = 1/5pF + 1/10pF
Ceq = 3.33pF

Charge and Voltage:

The charge on each capacitor is given by:

Q = CV

where Q is the charge, C is the capacitance, and V is the voltage across the capacitor.

The potential difference across each capacitor can be found using Kirchhoff's voltage law:

V = V1 + V2

where V is the voltage across the combination of capacitors, V1 is the voltage across the 5pF capacitor, and V2 is the voltage across the 10pF capacitor.

Ratio of Voltages:

We can find the ratio of V1 to V2 by using the fact that the charge on each capacitor is the same:

Q = CV1 = Ceq(V1 + V2)

Solving for V1, we get:

V1 = (Ceq/C)V - V2

Substituting in the values of Ceq, C, and V, we get:

V1/V2 = (1/2) - 1 = -1/2

Therefore, the potential difference across the 5pF capacitor is half of the potential difference across the 10pF capacitor, but in the opposite direction.

Answer

Ratio will be 6:3

Answer

Answer

Net capacitance will be 10/3so by  Q=CVwe get Q=10/3*9=30pCnow Q WILL BE SAME ACROSS both the capacitors bcz they r in series then get V1=Q/C1 AND V2=Q/C2

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