In the circuit in the figure, the capacitors are completely uncharged. The switc

Question

In the circuit in the figure, the capacitors are completely uncharged. The switch is then closed for a long time. As shown, R1 = 6.50 ?, R2 = 8.10 ?, 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 = 6.50 ?-resistor.

d) Find the potential difference across the R2 = 8.10 ?-resistor.

e) Find the potential difference across the 1.00 ?F-capacitor

In the circuit in the figure, the capacitors are completely uncharged. The switch is then closed for a long time. As shown, R1 = 6.50 ?, R2 = 8.10 ?, 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 = 6.50 ?-resistor. d) Find the potential difference across the R2 = 8.10 ?-resistor. e) Find the potential difference across the 1.00 ?F-capacitor

Explanation / Answer

a) After this has been closed for a long time, the capacitor will be as full as it's going to get. So the current through the 4 ohm resistor is zero.

b) There's no current through it, so there's no potential across it.

c) Essentially now the circuit consists of the battery and R1 and R2. Those resistors are in series so the total resistance is R1+R2. So the current flowing is 15.0 /(R1+R2).

So the potential across R1 is V1 = R1*15.0/(R1+R2)

V1 = 6.678 V

d) The potential difference across the R2 = 15 - 6.678 = 8.322 V

e) You just calculated the potential difference across R2. The 4 ?-resistor and the cap are in parallel with R2. So that potential is also across the 4 ?-resistor and the cap. Thereare zero volts across the 4 ?-resistor (see part b), so all of the potential you found to be across R2 is also across the cap.

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