please answer 59 and 60. Thanks! A 6.00 mu F capacitor that is initially uncharg

Question

please answer 59 and 60. Thanks! A 6.00 mu F capacitor that is initially uncharged is connected in series with a 4500 ohm resistor and a 500 V emf source with negligible internal resistance. Just after the circuit is completed, what are the voltage drop across the capacitor, the voltage drop across the resistor, the charge on the capacitor, and the current through the resistor? A long time after the circuit is completed (after many time constants), what are the values of the preceding four quantities? A capacitor is charged to a potential of 12.0 V and is then connected to a voltmeter having an internal resistance of 3.40 M ohm. After a time of 4.00 s the voltmeter reads 3.0 V. What are (a) the capacitance and (b) the time constant of the circuit? A 1.0 mu F capacitor is being charged by a 9.0 V battery through a 10 M Ohm resistor. Determine the potential across the capacitor at

Explanation / Answer

Just after the circuit is completed, the capacitor acts as a short circuit. So, the circuit will seem to anly have a resistor.

a)

So, voltage drop across capacitor = 0

b)

across resistor = V/R = 500 V

c)

Charge on the capacitor = 0

d)

Current through the resistor = V/R = 500/4500 = 0.11 A

e)

After a long time the capacitor charges completely and acts as an open circuit.

So, Voltage across capacitor = 500 V

Voltage across resistor = 0

Charge on the capacitor = C*V = 6*10^-6*500 = 3*10^-3 C

Current through resistor = 0

60)

Voltage across capacitor is given by:

Vc = Vo*e^(-t/RC)

where R = 3.4 Mohm = 3.4*10^6 ohm

and Vo = 12 V

Now, after 4s,

Vc = 3 V = 12*e^(-4/(3.4*10^6*C))

So, C = 8.5*10^-7 F <----------- capacitance

b)

Time constant = R*C = 3.4*10^6*8.5*10^-7 = 2.89 s

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