Question: In the circuit, in Fig. E26.47 the capacitors are initially uncharged, the battery has no internal resistance, and the ammeter is idealized. Find the ammeter reading (a) just after the switch S is closed and (b) after S has been closed for a very long time.

The potential across an uncharged capacitor is initially zero, so it behaves like a short circuit.

Here the switch is just closed which means that each capacitor acts as a short circuit. Hence the equivalent resistance for the circuit, for R₃ and R₅ are in parallel therefore the equivalent resistance is given by:

$R_{3,5}=\frac{R_3{R}_5}{R_3+R_5}$

Now, putting the values we get:

$$\begin{gathered} R_{3,5}=\frac{R_3{R}_5}{R_3+R_5} \\ =\frac{\left(50\Omega \right)\left(25\Omega \right)}{\left(50\Omega \right)+\left(25\Omega \right)} \\ =16.7\Omega \end{gathered}$$

Now the combination in series with and the equivalence resistance in the circuit is given by:

$$\begin{gathered} R_{eq}=R_{3,5}+R_2+R_6 \\ =16.7\Omega +75\Omega +15\Omega \\ =106.7\Omega \end{gathered}$$

Now using Ohms law, we get:

$$\begin{gathered} I=\frac{\epsilon }{R_{eq}} \\ =\frac{100V}{106.7\Omega } \\ =0.937A \end{gathered}$$

There the current just after the switch S is closed after the capacitors are initially uncharged is

A fully charged capacitor acts as an infinite resistance which blocks the current. When the capacitors block the currents all the resistors will be in series, hence the equivalent resistance in the circuit will be:

$$\begin{gathered} R_{eq}=15\Omega +25\Omega +25\Omega +75\Omega \\ =165\Omega \end{gathered}$$

According to Ohms law, we get the current in the ammeter:

$$\begin{gathered} I=\frac{\epsilon }{R_{eq}} \\ =\frac{100V}{165\Omega } \\ =0.606A \end{gathered}$$

Therefore, the current after the switch was closed for a long time is