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Wheatstone Bridge

In the adjoin...

Question

In the adjoining circuit diagram $E = 5 V, r = 1 W, R_{2} = 4 W, R_{1} = R_{3} = 1 W$ and $C = 3 m F$ . Then, the numerical value of the charge on each plate of the capacitance is

20 μ C

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12 μ C

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6 μ C

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3 μ C

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Solution

The correct option is C $6 μ C$
Since at $t = \infty$ capacitors behave as an open circuit therefore, no current will flow through $R_{1} a n d R_{3}$ .

Now Resistance of $R_{2} = 4 Ω$ and internal resistance of the circuit is $r = 1 Ω$

therefore, current throgh $R_{2}$ is $i = V / (R_{2} + r) = 5 / (4 + 1) = 1 A$

Now voltage across $R_{2}$ is $V_{2} = i R_{2} = 4 V$

Since, capacitors are connected in parallel to the R_2 therefore, voltage
across them is also $4 V o l t$

Therefore, potential across each capacitor will be $2 V o l t$

Therefore, charge on each capacitor $q = C V = 3 \times 10^{- 6} \times 2 = 6 μ C$

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Similar questions

E = 5 V, r = 1 Ω, R_{2} = 4 Ω, R_{1} = R_{3} = 1 Ω

C = 3 μ F

R_{1} = 10 Ω, R_{2} = 20 Ω, R_{3} = 40 Ω, R_{4} = 80 Ω

V_{A} = 5 V, V_{B} = 10 V, V_{C} = 20 V, V_{D} = 15 V

R_{1}

△ A B C

a < b < c

r

r_{1}, r_{2}, r_{3}

A, B, C

r < r_{1} < r_{2} < r_{3}

△ A B C, r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1} = \frac{r_{1} r_{2} r_{3}}{r}

C

E = 10

R_{1} = 2

R_{2} = 3

R_{3} = 6

L = 5

i_{1}

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