Consider a capacitor with a capacitance of lOuF. You connect it to a variable DC

Question

Consider a capacitor with a capacitance of lOuF. You connect it to a variable DC voltage source with a switch in the circuit and a resistor of 1000 ohm in series with the capacitor. (a) What is the relation between the steady-state voltage across the capacitor and the charge? (b) Before you close the switch, there is no charge on the capacitor. When you close the switch, current begins to flow from the voltage source at E volts. What is the relation between current and the voltage drop across the resistor in terms of the current and the resistance? (c) What is the voltage drop across the capacitor in terms of charge and capacitance? (d) Kirchoff s voltage law says that the voltage drop around any loop must be zero. Write the equation for the voltage drop across the resistor, capacitor and voltage source. (e) Solve the equation in part D to derive the time course of charging of the capacitor. (f) Solve the equation in part C to derive the time course of the current.

Explanation / Answer

2. given capacitance, C = 10 micro F

resistor vaslue, R = 1000 ohms

a. steady state voltage across capacitor = V

but Q = CV [ where Q is charge on capacitor, C is capacitance]

Q = 10^(-6) *V

b. if the voltage source i E volts

then

E = iR + q/C [ i is current through the circuit and q is charge on capacitor]

so voltage drop across resistor = V

then V = iR [ i is current rthrough the resistor]

c. voltage drop across capacitor, Vc = q/C [ where q is charge on capacitor and C is its capacitance]

d. so according to kirchoff's law

E = iR + q/C

e. but i = dq/dt

so E = Rdq/dt + q/c

(E - q/C) = Rdq/dt

dt/R = dq/(E - q/C)

integrating

t/R = -Cln(E - q/C)

-t/RC = ln(E- q/C) + k [ where k is constant of integrations]

at t= 0, q = 0

0 = ln(E) + k

k = -ln(E)

so, -t/RC = ln(E - q/c) - ln(E) = ln(1 - q/EC)

1 - q/EC = e^(-t/RC)

q = EC(1 - e^(-t/RC))

so when t -> infinity

maximum charge, qo = EC

f. dq/dt = i

i = EC*e^(-t/RC)/RC = Ee^(-t/RC)/R

at t = 0, i = E/R ( maximum current)

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