In the circuit in the figure (http://s19.postimage.org/sp4vsbjvn/P054figure.png)
ID: 2154580 • Letter: I
Question
In the circuit in the figure (http://s19.postimage.org/sp4vsbjvn/P054figure.png), the capacitors are completely uncharged. The switch is then closed for a long time. As shown, R1 = 6.20 ?, R2 = 8.40 ?, and V = 15.0 V.a) Calculate the current through the 4.00 ?-resistor.
b) Find the potential difference across the 4.00 ?-resistor.
c) Find the potential difference across the R1 = 6.20 ?-resistor.
d) Find the potential difference across the R2 = 8.40 ?-resistor.
e) Find the potential difference across the 1.00 ?F-capacitor
Explanation / Answer
if we are taken da values supoose R1 = 6.50 ?, R2 = 8.50 ?, and V = 14.0 V. then answer is a and b are correct. c) The other two resistors are in series. The voltage drop across one of them is E_r1 = R1*E/(R1 + R2) E_r1 = 6.5 * 14 / (6.5 + 8.5) = 6.07 volts d) E_r2 = 8.5 * 14 / (15) = 7.93 volts e) The capacitor when fully charged has an extremely high resistance. You can run it through with a resistance of (say 100,000 ohms if you like. What you will find is that you can treat it as an open circuit. The point is that the voltage drop will be the same as for R2. Whatever R2 sees, the capacitor 4 ohm resistor will see. The answer is 7.93 V