Question

Calculate the currents i, $i_1$ and $i_2$ in the given Wheatstone's network of resistors, The internal resistance of the 6 V battery is $0.4\Omega$.

Answer 1

due to wheatstone's network.

current through G is zero circuits will look like as shown.

total resistance is $\frac{9\times 6}{9+6}\Omega +.4\Omega =3.6\Omega +.4\Omega =4\Omega$.

current $i=\frac{6V}{4\Omega }=1.5A$

current $i_1$bu using current division

$i_1=\frac{9}{9+6}\times i=\frac{9}{15}\times 1.5=.9A$

similiarly, $i_2=\frac{6}{6+9}\times i=\frac{6}{15}\times 1.5=.6A$