RC circuit problem

To calculate the current in the circuit in the case when the switch is open and steady state is reached, I assumed that no current will flow "across" the capacitors and so the current I will simply be emf/(r+R1+R2) = 12 V/(1+3+2) = 2 A.

The voltage drop across r would be V=IR=2 A * 1 ohms = 2 V, therefore the voltage across the capacitors would be 10 V.

Since C1 and C2 are in series, Q1=Q2=CV where C is the combined capacitance equal to (4/3)F and V=10V.

Hello.

Correct.




Correct.





You need to be careful here.

For two series capacitors on their own (i.e. no parallel resistors), the calculation for the equivalent capacitance is correct (i.e. switch open):

$\frac{1}{C_{tot}} = \frac{1}{C1} + \frac{1}{C2} = \frac{1}{2} + \frac{1}{4} = \frac{3}{4}$

$C_{tot} = \frac{4}{3} \mathrm {F}$



But when the switch is closed, the presence of those resistors change things:

We are still dealing with steady state conditions (i.e. at t>>0), but the capacitors will be forced into a new steady state condition. When achieved, the new steady state means the only current flowing will also be via the resistors as you have previously correctly deduced, because the capacitors will still achieve a new full charge and therefore no further current flows into them.

Notice however, in both open and closed switch conditions, there will be a potential difference developed across each resistor.

i.e. the resistors form a potential divider, So you need to work out the potential difference across each resistor first.

The closed switch forces those potential divider p.d.'s to be developed in parallel with their associated capacitor. And since the p.d's will be different for each parallel RC combination, the capacitors must achieve a new steady state.

Now it's a case of using the appropriate resistor p.d. and Q=CV to work out the charge on each capacitor.

Notice that the final charge is different to the isolated series case and no rules are violated. i.e. 10V will still be placed across the series combination, steady state current only flows via the resistors series path, p.d.'s across the resistors obey Ohms and Kirchoff's laws.

Thanks for the reply. The p.d. across R1 would be 6V and the p.d. across R2 would be 4V, so the associated drops for C1 and C2 would also be 6V and 4V. Therefore Q1=2F*6V = 12C and Q2=4F*4V=16C?