Question

In the following circuit, the current through the resistor $R(=2\Omega )$ is $I$ Amperes. The value of $I$ is

Answer 1

Due to balanced wheatstone bridge ABCEA $BE=8\Omega$ can be taken out.

$R_{AC}=\frac{(1+2)(2+4)}{(1+2)+(2+4)}=\frac{18}{9}=2$

Now, ACFGHA is a balanced Wheatstone bridge, hence $HC=10\Omega$ can be taken out.


$R_{AG}=\frac{6\ast 18}{6+18}=\frac{108}{24}=4.5\Omega$

$R_{total}=2+4.5=6.5\Omega$
Hence, current $i=\frac{V}{R_{total}}=\frac{6.5}{6.5}=1A$