Two resistors, R 1 = 9.00Ohm and R 2 = 4.50Ohm and two initially uncharged capac

Question

Two resistors, R1 = 9.00Ohm and R2 = 4.50Ohm and two initially uncharged capacitors, C1 = 0.520uF and C2 = 0.260uF are arranged as shown in the figure below.

With a potential difference of 26.4V across the combination, what is the potential at point a with S open? (Let V = 0 at the negative terminal of the source.)

What is the potential at point b with the switch open?

When the switch is closed, what is the final potential of point b?

Two resistors, R1 = 9.00Ohm and R2 = 4.50Ohm and two initially uncharged capacitors, C1 = 0.520uF and C2 = 0.260uF are arranged as shown in the figure below. With a potential difference of 26.4V across the combination, what is the potential at point a with S open? (Let V = 0 at the negative terminal of the source.) What is the potential at point b with the switch open? When the switch is closed, what is the final potential of point b?

Explanation / Answer

Resistance R1 and R2 are in series , so total resistance = R1 + R2

After a long time the two capacitors are fully charged and all the current flow through the resistance branch which can be given as ::

i = V/ ( R1 + R2)

V = 26.4 volts   ,   R1 = 9 ohm , R2 = 4.5 ohm

So i = 26.4 volts / (9 ohm + 4.5 ohm)

i = 1.96 A

So Voltage at ''a'' is = Voltage across R2 = i R2 = (1.96 A) (4.5 ohm) = 8.82 volts

b) C1 and C2 are in series and their combination is given as

C12 = C1 * C2 / (C1 + C2)

Voltage across the combination = V

So Charge Flowing in Capacitor = Q = C12 *V

Q = C1 * C2 * V / (C1 + C2)

Voltage at b is Voltage across C2 which is given as

V2 = Q/C2

V2 = C1 * V / (C1 + C2) = (0.520 x 10-12) (26.4)/((0.520 x 10-12 +0.260 x 10-12) =17.6 volts

V2 = 17.6 volts

c) 0 volts. Since no current flow in capacitors

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