Circuit Analysis: Finding Current, Charge, and Voltage with a Flash Unit

[~christina~]

Homework Statement

A photographer's electronic flash unit consists of a 150.0-kΩ resistor in series with a 22-µF capacitor and a 12.0 V source of emf. The flashbulb is placed in parallel with the capacitor so that when a sufficient charge is stored on the capacitor switch S2 can be closed and the capacitor quickly discharges through the small resistance of the bulb, causing the flash. Suppose switch S2 remains open and at t = 0 s switch Sl is closed.

(a) At what time after the capacitor begins to charge has the current decreased to one-half its initial value?

(b) What is the charge on the capacitor at this time?

(c) What is the voltage across the capacitor at this time?

(d) Sketch graphs of the current through this circuit and the charge on the capacitor as functions of time (two graphs

http://img366.imageshack.us/img366/695/picture6ni5.th.jpg

Homework Equations

\tau= RC

I_i= \frac{\epsilon} {R}
I(t)= \frac{\epsilon} {R} e^{\frac {-t} {RC}}

The Attempt at a Solution

a) At what time after the capacitor begins to charge has the current decreased to one-half its initial value?

\tau= RC= (22x10^{-6} F)(1.50x10^5 \omega )= 3.3s

max current: I_i= \frac{\epsilon} {R} = 12.0V/ 1.50x10^5 \omega= 8x10^{-5} A

current as a function of time: I(t)= \frac{\epsilon} {R} e^{\frac {-t} {RC}}

half of the total current: 4x10^{-5}A

I think I would find the time this way but I'm not sure if it is right.
time it takes for the current to half
I(t)= (8x10^-5A)\frac e^{\frac {-t} {3.3s}}
0.5A= \frac e^{\frac {-t} {3.3s}}
ln 0.5A= \frac{-t} {3.3s}}
t= 2.287s

(b) What is the charge on the capacitor at this time?

q(t)= C \epsilon (1-e^{-\frac{t} {RC}})
C \epsilon = (22x10^{-6}F)(12.0V)= 2.64x10^{-4}C
q(t)= (2.64x10^{-4}C)(1-e^{-\frac{2.287s} {3.3s}})
q(t)= (2.64 x10^ {-4}C)(0.499941)=1.319844x10^{-4}C

(c) What is the voltage across the capacitor at this time?

Not sure how to find this..

(d) Sketch graphs of the current through this circuit and the charge on the capacitor as functions of time (two graphs

wouldn't this look the typical graph of

charge vs time (exponential curve going up)=> |/|
current vs time (exponential curve going down)=> |\|

I'd appreciate it if someone could help me out with part c and also checking whether I did the other parts correctly.

Thanks a lot

 

[kamerling]

part a. the method and the result is Ok. 0.5 shouldn't have a u unit of amperes as this is a dimensionless number.

part b. Ok. A shorter method would be to notice that since the current is now half the initial current, half of the voltage of the voltage source is the drop across the resitor.

part c. you already gave q(t) in part b. Q = CV

part d. you already computed I(t) and Q(t)

 

[~christina~]

part a. the method and the result is Ok. 0.5 shouldn't have a u unit of amperes as this is a dimensionless number.

ok

part b. Ok. A shorter method would be to notice that since the current is now half the initial current, half of the voltage of the voltage source is the drop across the resitor.

yep, didn't notice that until you mentioned it.

part c. you already gave q(t) in part b. Q = CV

oh...so I'd just find the voltage by using the charge I found for the specific time.

part d. you already computed I(t) and Q(t)

okay.


Thanks for your help kamerling

 

[!MAD DAWG!]

ya thanks