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Electrostatic Potential and Capacitance

Capacitive circuit

Analyze the given circuit in the steady state condition. Charg

Question

Hard

Solving time: 4 mins

Analyze the given circuit in the steady state condition. Charge on the capacitor is $q_{0} = 16 μ C$ .

(a) Find the current in each branch
(b) Find the emf of the battery.
(c) If now the battery is removed and the points $A$ and $C$ are shorted. Find the time during which charge on the capacitor becomes $8 μ C$ .

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Book:

Electricity and Magnetism (DC Pandey)

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Daniel Discussed

Analyze the given circuit in the steady state condition. Charge on the capacitor is

q_{0} = 16 μ C

(a) Find the current in each branch
(b) Find the emf of the battery.
(c) If now the battery is removed and the points

A

and

C

are shorted. Find the time during which charge on the capacitor becomes

8 μ C

11 mins ago

Discuss this question LIVE

11 mins ago

Text solutionVerified

(a) Current in lower branch

= E /8 = 3 A

Current in upper branch

= E /9 = 24/9 = 2.67 A

(b)

PD

across the capacitor

= E /2 - E /3 = E /6

From

q = C V

, we have

16 = (4) \frac{E}{6}

∴ E = 24 V

(c) After short-circuiting the battery, we will have to find net resistance across capacitor to calculate equivalent value of

τ_{C}

in discharging.

3Ω

and

6Ω

are in parallel. Similarly,

4Ω

and

4Ω

are in parallel. They are then in series.

∴ R_{n e t} = 4Ω,

τ_{C} = C R_{net} = (4 \times 4) μ s = 16 μ s

During discharging

q = q_{0} e^{- t / τ_{C}}

8 = 16 e^{- t /16}

Solving this equation, we get

t = 11.1 μ s

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Question Text	Analyze the given circuit in the steady state condition. Charge on the capacitor is $q_{0} = 16 μ C$ . (a) Find the current in each branch (b) Find the emf of the battery. (c) If now the battery is removed and the points $A$ and $C$ are shorted. Find the time during which charge on the capacitor becomes $8 μ C$ .
Topic	Electrostatic Potential and Capacitance
Subject	Physics
Class	Class 12
Answer Type	Text solution:1
Upvotes	51

Text solutionVerified

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