course phy 232
i worked throught this problem and think it could possibly be correct. ill explain what i did.????????????
Problem Number 9
A generator has an internal resistance of 42 ohms, creates a potential difference of 4.2 volts. The generator is connected to an uncharged capacitor with capacitance .62 Farad in series with a bulb whose resistance is 49 Ohms. Approximately how long will it take to charge the capacitor to .1 * 4.2 volts? How much charge will there be on the capacitor at this time? What will be the voltage across the bulb at this time?
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so i have I coming out of the generator = 4.2 V / 42 ohm = .1 C/s this is convenient with the question wondering how long to charge the capacitator.
for the capacitator i got Q= C*(IR) =.62 farads*.1 C/s*49ohm = 3.04 C
the time to reach .1*4.2 volts is 4.2 seconds becuase the current i think is the same as the .1 from the generator. unless im needing to find how much current remains on the other side of the capacitator, but that doesnt make sense really. i notice that the value above represents units for power generated.
now i need to find the charge at t = 4.2 second
I want to multiply the current by the time 4.2 seconds, because the units are c/s. it seems like maybe something i could do.
this leaves 3.04C*4.2s = 12.8 volts seconds...
doing this for the voltage yields 49ohms*.1C/s*4.2seconds = 20.6
im not sure this is the right way to go about this.
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The resistance of the circuit is 42 ohms + 49 ohms = 91 ohms, not 42 ohms. This would change the result somewhat.
If the resistance of the circuit was 42 ohms, then the current would be .1 C / s for the reasons you give. We'll use this value:
At .1 * 4.2 volts = .42 volts the charge on a .62 Farad capacitor is Q = C * V = .62 farads * .42 volts = .62 C / volt * .42 volts = .26 coulombs.
If the current remains .1 C / s it will take 2.6 seconds to build the .26 coulomb charge.
The current does not remain .1 C / s; the capacitor ends up with a voltage of .42 volts. This voltage opposes that of the generator, reducing the current. However there is only a 10% reduction in the 4.2 volt potential difference, hence a 10% reduction in the current at the final instante when the specified charge has been transferred. There is thus an average of only about a 5% reduction in the current and a 5% increase in the time required to build the charge. So the 2.6 second estimate is not much different from the actual time required.
The voltage at any instant is (V_gen - V_cap) so the current is (V_gen - V_cap) / R, where V_gen is the generator voltage and V_cap the capacitor voltage, R the resistance of the circuit.
V_cap = Q / C, where Q is the charge on the capacitor.
The current is the rate at which charge is stored on the capacitor.
Thus
I = (V_gen - V_cap) / R becomes
dQ/dt = (V_gen - Q / C) / R, which can be written
dQ / dt = (V_gen / R - Q / (R C)). V_gen / R is just the initial current I_0 so this can be written
dQ / dt = (I_0 - Q / (R C) ).
This equation is solved for Q as a function of clock time t, and results in a charge function of the form
Q(t) = V_gen * C * (1 - e^(-t / (R C) ) ).
That is, the charge starts at zero and approaches V_gen * C as an asymptote.
This can be solved for Q = .1 * V_gen * C:
.1 * V_gen * C = V_gen * C * (1 - e^(-t / (R C) )) so
(1 - e^(-t / (R C) )) = .1 so that
e^(-t / (R C) ) = .9 and
-t / (R C) = ln(.9). Thus
t = R C ln(.9) = -42 ohms * .62 Farads * (-.105) = 2.7 seconds.
If you work it out exactly you'll see that it takes just about 5% longer than the previous result.