Question

Four resistors of $12\Omega$ each are connected in parallel. Three such combinations are then connected in series. What is the total resistance ? If a battery of $9V$ emf and negligible internal resistance is connected across the network of resistors, fond the current flowing through each resistor.

Answer 1

Four resistance of $12\Omega$ of each are connected in parallel.
Then,
$\frac{1}{R_{||}}=\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}$

$=\frac{4}{12}=\frac{1}{3}$

$R_{||}=3\Omega$.

Three such combinations are connected is series. Then,
$R_{eq}=R_{||}+R_{||}+R_{||}$
$=3+3+3$
$=9\Omega$
$V=9V$

$I=\frac{V}{R_{eq}}=\frac{9}{9}=1A$

As all resistors have same resistance
$i_1=i_2=i_3=i_4$
Also, $i_1+i_2+i_3+i_4=I$
$4i_1=1$
$i_1=0.25$
For each capacitor $i=0.25A$