Question

In the circuit shown, r=4Ω,C=2μFr=4Ω,C=2μF
(a) Find the current coming out of the battery just after the switch is closed.
(b) Find charge on each capacitor in the steady condition.

Answer 1

Due to symmetry, the points of equal potential are joined together, and the circuit may be reduced as
$R_{13}=\frac{r}{2}+\frac{r}{4}+\frac{r}{4}+\frac{r}{2}=\frac{3}{2}r$
Here $r=4\Omega ,R_{13}=\frac{3}{2}\left(4\right)=6\Omega$
Thus
$i=\frac{24}{R_{13}}=\frac{24}{6}=4A$
(b) In the steady state condition, the circuit may be reduced as $R_{13}=2r=2\left(4\right)\Omega =8\Omega$
$I=\frac{24}{8}=3A$
$V_{56}=\left(2r\right)\frac{I}{2}=Ir=\left(3\right)\left(4\right)=12V$
Equivalent capacitance between 5 and 6 is
$C_{56}=\frac{C}{2}=1\mu F$
$\therefore q_{58}=C_{56}{V}_{56}=12\mu C$
Now, $V_{97}=V_{17}-V_{19}=-r\frac{I}{2}+2\frac{rI}{2}=\frac{rI}{2}=\frac{\left(4\right)\left(3\right)}{2}=6V$
$\therefore q_{97}=C{V}_{97}=\left(2\right)\left(6\right)=12\mu C$
similarly $q_{89}=12\mu C$