A) What is the charge on the capacitor after the switch has been closed for t=2.

Question

A) What is the charge on the capacitor after the switch has been closed for t=2.95x10-2 seconds?

B) What is the current through the circuit after the switch has been closed for t=2.02x10-2 seconds?

C)What is the voltage across the capacitor after the switch has been closed for t= 2.02x10-2 seconds ?

The resistance in the circuit is R= 534Ohm the capacitor has capacitance C= 48.0 microFarads and the battery maintains the emf of E= 14.0V. The switch is closed at time t=0s, which causes the capacitor to begin to charge. What is the charge on the capacitor after the switch has been closed for t=2.95 Times 10-2 seconds? What is the current through the circuit after the switch has been closed for t=2.02 Times 10-2 seconds? What is the voltage across the capacitor after the switch has been closed for t= 2.02 Times 10-2 seconds ?

Explanation / Answer

we have the formula,

Vc=Vo(1-e^(-t/T))

T= time const=RC=534*48*10^-6

=0.25632 sec sec

so,

A)

at t=2.95*10^-2 sec

Vc=14*(1-e^(-2.95*10^-2/0.25632)

=1.52208 V

chrage q = CV

=48*10^-6*1.52208

=73.05984 *10^-6 coulumbs

=0.7305984 milli coulumbs........................

B)

at t=2.02*10^-2 sec

q = q0(1-e-t/T)

q = 0.73*10^-3(1-e-2.02*10^-2/0.25632)

q = 0.73*10^-3(1-0.5)

= 0.055334*10^-3 C

current i = dq/dt

= (q0-q)/T

=(0.73-0.055)*10^-3/0.25632

=2.633 milliampears...................

C)

at t=2.02*10^-2 sec

q=0.05533*10^-3 C

so,

vC=q/c

=0.05533*10^-3/48*10^-6

= 1.1527 V .......................

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