(25%) Problem 4: A bank of batteries, total emf 4.5 V, is connected with resisto
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(25%) Problem 4: A bank of batteries, total emf 4.5 V, is connected with resistor R-120 kQ, capacitor C= 680 pF. and two-pole switch S as shown. The switch is initially set to point b so that the resistor and capacitor are in series. The switch is left in this position for a sufficiently long time so that the capacitor is completely discharged Randomized Variables =4.5 V R= 120k C- 680 pF theespertta.com 25% Part (a) The switch is disconnected from point b and connected to point a. What will happen to the voltage across the capacitor? Grade Summary Deductions 0% Potential The voltage will immediately rise to The voltage will not change The voltage will immediately drop to zero. The voltage will increase exponentially with time 100% Amenpts remaining OThe voltage will drop exponentially with time OThe voltage will drop linearly with time OThe voltage will increase linearly with time per attempt detailed vew l give Hints: Feedback: 25% Part (b) Calculate the time constant, t(in seconds), for this circuit - 25% Part (c) Determine an expression for the charge on the capacitor as a function of time, q(t), in terms of parameters given in the original problem statementExplanation / Answer
Here ,
E= 4.5 V
R = 120 Ohm
C = 680 pF
part A)
the voltage will increase exponentially with time
part b)
Time constant = R * C
Time constant = 120 *10^3 680 *10^-12 s
Time constant = 8.16 *10^-5 s
part C)
Now , for the charge on the capacitor
q = qo * (1 - e^(-t/T))
q = 4.5 * 680 *10^-12 * (1 - e^(-t( 8.16 *10^-5 )))
q(t) = 3.06 *10^-9 * (1 - e^(-t( 8.16 *10^-5 )))
part d)
at time , t = 190 us
v(t) = vo * (1 - e^(-t/T))
v(t) = 4.5 * (1 - e^(-t( 8.16 *10^-5 )))
v(t) = 4.5 * (1 - e^(-190 *10^-6/(8.16 *10^-5)))
v(t) = 4.06 V
the voltage drop across the capacitor is 4.06 V
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