A circuit is connected as shown in the figure with the switch open. When the switch is closed, what is the total amount of charge (in μC ) that flows from Y to X ?

Question

A circuit is connected as shown in the figure with the switch open. When the switch is closed, what is the total amount of charge (in μC ) that flows from Y to X ? <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-rzwcs7hpc5moh0yh.jpg" /> (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

When switch is open capacitors are in series and net potential difference across them is

9 v

Charge on each capacitor before closing the switch is

q = C_{e q} V = \frac{3 \times 6}{3 + 6} \times 9 = 18 μ c

After closing the switch current of

I = \frac{V}{3 + 6} = \frac{9}{9} = 1 A

will flow across the resistors .potential difference across resistors and capacitors are

V_{3 o hm} = I R = 1 \times 3 = 3 V V_{3 o hm} = V_{3 μ F} = 3 V V_{6 o hm} = I R = 1 \times 6 = 6 V V_{6 o hm} = V_{6 μ F} = 6 V

Charge on each capacitor after closing the switch will be

q_{3 μ F} = C V = 3 \times 3 = 9 μ c q_{6 μ F} = C V = 6 \times 6 = 36 μ c

Hence, charge flown from

Y t o X

is

q_{f l o w} = (36 - 18) + (18 - 9) = 27 μ C

Q. A circuit is connected as shown in the figure with the switch open. When the switch is closed, what is the total amount of charge (in μC ) that flows from Y to X ?

Q. A circuit is connected as shown in the figure with the switch open. When the switch is closed, what is the total amount of charge (in $μ C$ ) that flows from $Y$ to $X$ ?