# RC circuit confusion [closed]

I was doing some RC problems in A levels, and in ine of thr problems it was about a circuit attached as a picture:

My question is can someone clearly explain to me why when time goes to infinity , current through resistor in R3 goes to zero? But when time = 0 current through resistor in R3 is not zero?

My other question is why when time is infinity current through resistor R3 is zero but current though resistor in R2 is not zero?

I would appreciate the help thank you.

## closed as off-topic by Jon Custer, ZeroTheHero, John Rennie, Kyle Kanos, BuzzJan 11 at 17:11

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• Yes my bad thats what I was trying to say, could you explain please? Also why is current through R2 is not zero as time goes to infinity? – Aurora Borealis Jan 11 at 2:52

Look at the branch of the circuit that has $$C$$ and $$R_3$$. The top of $$R_3$$ and the bottom of $$C$$ will experience some potential difference. Therefore, we can treat that part of the circuit as a charging RC circuit once the switch is closed.

We know that for an RC curcuit $$I(t)=\frac VRe^{-t/RC}$$ So after a long time the current drops to $$0$$ because the capacitor becomes fully charged, but at $$t=0$$ the current is at its maximum value of $$V/R$$ (as if the capacitor is just a wire for this instant since there is no charge on it).

After a long time, since there is no current flowing through the capacitor branch, you can essentially ignore that part of the circuit. Then the circuit reduces to just two resistors ($$R_1$$ and $$R_2$$) in series with a battery.

• thank you for your response, my confusion now is isnt the equation you used, isnt that for the discharging capacitor? if so how is this problem suddenly become the discharging case? – Aurora Borealis Jan 11 at 3:21
• @AuroraBorealis That equation works for both. In the charging case $V$ is the potential of the battery charging the capacitor. In the discharging case $V$ is the initial potential difference across the capacitor. In either case the current starts at a maximum and exponentially decays to $0$ with time constant $RC$ – Aaron Stevens Jan 11 at 3:24
• Then when do we use the equation $V(t)=\epsilon (1-e^{-t/RC})?$ – Aurora Borealis Jan 11 at 3:28
• @AlfredCentauri I agree. I have edited the answer. Thanks. – Aaron Stevens Jan 11 at 3:43
• @AuroraBorealis The comments of an answer is not a place for me to teach/you to learn about RC circuits. If you have another question please post a new question. – Aaron Stevens Jan 11 at 3:45

My question is can someone clearly explain to me why when time goes to infinity , current through resistor in R3 goes to zero?

Sure. As time goes to infinity, each circuit voltage and current variable asymptotically approaches a constant value (stop here and think about this for a bit if it isn't clear). This 'final' state is called DC steady state.

Now, recalling that all the voltages and currents are constant at this 'final' time, think about what this implies for the capacitor current which is given by

$$i_C = C\frac{dv_C}{dt}$$

Since the capacitor current is proportional to the rate of change of the capacitor voltage, and since the capacitor voltage is constant (zero rate of change) in DC steady state, the capacitor current is zero in DC steady state.

Finally, since $$R_3$$ is in series with the capacitor, the current through $$R_3$$ is identical to the the current through the capacitor.

Thus, conclude that as time goes to infinity, the current through $$R_3$$ goes to zero.

My other question is why when time is infinity current through resistor R3 is zero but current though resistor in R2 is not zero?

Why would it be zero? In DC steady state, there is zero current through $$R_3$$ but that doesn't imply there is zero volts across $$R_2$$. Only if there is zero volts across $$R_2$$ can there be zero current through $$R_2$$.