In the adjoining circuit diagram, E=5 V , r=1 Ω , R2=4 Ω , R1=R3=1 Ω and C=3 mF . Then the numerical value of the charge on each plate of the capacitor is

Question

In the adjoining circuit diagram, E=5 V , r=1 Ω , R2=4 Ω , R1=R3=1 Ω and C=3 mF . Then the numerical value of the charge on each plate of the capacitor is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-db3rf75tznxr.png" /> (A) 24μC (B) 12μC (C) 6 μC (D) 3μC

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-db3rf75t0bak.png" /> Current through resistors R1 and R3 in steady state is zero. Therefore current through R2 is i2=(E/R2 + r)=(5/4 + 1)=1A Therefore charge on capacitor is q=C((i2 R2/2))=6 Î¼C

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

$24 μ C$

$12 μ C$

$6 μ C$

$3 μ C$

Current through resistors $R_{1}$ and $R_{3}$ in steady state is zero.
Therefore current through $R_{2}$ is
$i_{2} = \frac{E}{R _{2} + r} = \frac{5}{4 + 1} = 1 A$
Therefore charge on capacitor is
$q = C (\frac{i _{2} R _{2}}{2}) = 6 μ C$

Q. In the adjoining circuit diagram, E=5V , r=1Ω , R2​=4Ω , R1​=R3​=1Ω and C=3mF . Then the numerical value of the charge on each plate of the capacitor is

24μC

12μC

6μC

3μC