in the circuit shown (http://www.freeimagehosting.net/1t127
) assume the battery emf is 35.0 V, R = 2.40 MΩ;, and C = 5.59 µF. The switch is closed at t = 0. At what time t will the voltage across the capacitor be 23.6 V
RC = 2.40 M x 5.59 µ = 13.4 s
voltage on a cap, charging
v = v₀[1–e^(–t/RC)]
23.6 = 35[1–e^(–t/13.4)]
1–e^(–t/1.87) = 0.674
e^(–t/1.87) = 0.352
–t/1.87 = 0.326
t = 4.4 sec
But it's wrong. Any help is highly appreciated. Thanks so much!
2) This question is an easy but im not sure where im wrong.
Find the unknown emf and the unknown resistor in the circuit. I found the R but couldn't find the emf. I thought this mechanism would work to find the emf, but it didn't: 10.4 + 1.65 = 12.05
4.00 * 12.05 = 48.2
The picture is here: http://www.freeimagehosting.net/fgg2f
Any help is SO much appreciated.
1) How did the time constant tc magically change from 13.4 seconds to 1.87 seconds? Putting the proper time constant 13.4sec yields a time of 15.05 seconds.
2) the voltage you got, 48.2volts, is the voltage across the 4ohm resistor, but that is NOT the voltage of the unknown EMF, V1. To find that you will need to use Kirchoff's Loop Rule [a statement of energy conservation] which says that the sum of the EMF's around any closed conducting path must be equal to the sum of the voltage drops around that same path. In this case:
Solving for V1, the unknown voltage: