BR electric current problem #12

[Bumbl3b33]

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The question states: ...the capacitor becomes fully charged. the charge on the capacitor can now be written as...

I thought that when the capacitor becomes fully charged it has the same voltage drop as the EMF source?

 

[Pisiform]

The question states: ...the capacitor becomes fully charged. the charge on the capacitor can now be written as...

I thought that when the capacitor becomes fully charged it has the same voltage drop as the EMF source?

You would be correct if we only had a resistor R3 and capacitor in parallel, no third R1 resistor in series. The presence of R1 changes the solution. When the circuit has been closed for a long time, the capacitor is fully charged and so no current flows through capacitor part of the circuit and since no current flows .. this makes the top loop an open circuit. Because it is an open circuit now, you can ignore the capacitor loop. So you are only left with a battery and two resistors in series. Find out the total current I = V/R => I = EMF/R1+R3. You can find the voltage across the R3 resistor V = IR => (EMF/R1+R3) * R3. And since R3 is in parallel with the capacitor the voltage drop across the resistor would be same as the voltage drop in the capacitor, you can find
Q = C * (EMF/R1+R3) * R3