The circuit in Figure P28.75 contains two resistors, R_1 = 1.00 k Ohm and R_2 =

Question

The circuit in Figure P28.75 contains two resistors, R_1 = 1.00 k Ohm and R_2 = 3.00 k Ohm, and two capacitors, C_1 = 2.00 A mu F and C_2 = 4.00 A mu F, connected to a battery with emf epsilon = 130 V. No charge is on either capacitor before switch S is closed. Determine the charges q_1 and q_2 on capacitors C_1 and C_2, respectively, after the switch is closed. (Suggestion: First reconstruct the circuit so that it becomes a simple RC circuit containing a single resistor and single capacitor in series, connected to the battery, and then determine the total charge q stored in the equivalent circuit.)

Explanation / Answer

given

R1 = 1000 Ohm ,

R2 = 3000 Ohm

C1 = 2 X 10-6 F ,

C2 = 4 X 10-6 F

e = 130 volts

R = 1 / ( 1 / R1 + 1 / R2 )

R = 750 Ohm

C = C1 + C2

C = 6 X 10-6 F

t = R1 C1

t = 1000 X 2 X 10-6

t = 0.002 sec

and q2 = ( C2 / C1 ) q1

q2 = ( 4 / 2 ) X q1

q2 = 2 q1

q1 + 2 q1 = Q ( 1 - exp(t / 0.002 ) )

q1 = 0.33 Q ( 1 - exp(t / 0.002 ) ) ----- 1

q2 = 2 ( 0.33 Q ( 1 - exp(t / 0.002 ) ) ) ------ 2

q2 = 0.66 Q ( 1 - exp(t / 0.002 ) )

Q = 6 X 10-6 X 130

Q = 780 X 10-6 C

q1 = 0.33 Q ( 1 - exp(t / 0.002 ) )

substituting Q is in eq 1

q1 = 0.33 X 780 X 10-6 ( 1 - exp(t / 0.002 ) )

q1 = 257.4 X 10-6 ( 1 - exp(t / 0.002 ) )  

q2 = 0.66 Q ( 1 - exp(t / 0.002 ) )

substituting Q is in eq 2

q2 = 0.66 X 780 X 10-6 ( 1 - exp(t / 0.002 ) )

q2 = 514.8 X 10-6 ( 1 - exp(t / 0.002 ) )  

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