Question

The network shown in the figure is assembled with uncharged capacitors X, Y, and Z, with , and and open switches, S1 and S2. A potential difference Vab = +120 V is applied between points a and b. After the network is assembled, switch S1 is closed for a long time, but switch S2 is kept open. Then switch S1 is opened and switch S2 is closed. What is the final voltage across capacitor X?

Answer 1

Complete Question

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Answer:

The final voltage across X is  $V_X = 94 V$

Explanation:

First we are going to obtain the equivalent capacitance for the whole circuit

to do this we will first find the capacitance of the capacitors connected in series

This can be mathematically evaluated as

            $\frac{1}{C_s} = \frac{1}{C_Y} + \frac{1}{C_Z}$

Substituting values

           $C_s = \frac{3 \mu F * 3 \mu F }{(3 \mu F ) + (3 \mu F)}$

           $C_s = \frac{9 \mu F ^2}{6 \mu F}$

            $C_s =1.5 \mu F$

Now the equivalent capacitance of the capacitors connected in series  $C_s$ which is connected in parallel to $C_X$

So the equivalent capacitance of the whole circuit is mathematically evaluated as

        $C_T = C_X + C_s$

          Substituting values

      $C_T = 1.5 + 4$

        $C_T = 5.5 \mu F$

Now the total charge stores in these capacitors is mathematically evaluated as

             $Q = C_{T} V_{ab}$

Substituting values

             $Q = 5.5 \mu * 120$

             $Q = 660 \mu C$

When Switch $S_2$ is closed the capacitor $C_Y$ do not current anymore which implies that it has been excluded from the circuit

 Hence the equivalent capacitance becomes

          $C_T__{1}} = C_Z + C_X$

           $C_T__{1}} = 3 \mu F + 4 \mu F$

           $C_T__{1}} =7 \mu F$

Now at the point $S_2$  is closed the Voltage across the capacitor $C_X$ sis mathematically represented as

           $V_X = \frac{Q}{C_T__{1}}}$

            $V_X = \frac{600 \mu C}{7 \mu F}$

            $V_X = 94.2 V$

             $V_X = 94 V$