Physics - Current Electricity Question with Solution | TestHub

As can be seen from the first circuit above, $CB\&\text{A}B$ are in series which is in parallel with $C\text{A}$. So, the net resistance is

$R^{\prime }=\frac{6}{2}\Omega =3\Omega$ as shown in the second diagram.

From the second diagram, $D\mathrm{C}\&\mathrm{C}\text{A}$ is in series which is parallel to $D\text{A}$. Hence the resistance is

R ' ' = \left(\frac{6 \times 3}{(6 + 3 )}\right) \Omega = 2 \Omega as shown in third diagram.

From third diagram, both $ED\&D\text{A}$ are in series which is parallel to $E\text{A}$.

The equivalent resistance as can be seen from the above simplified circuit is $R_{eq}=2\Omega$.

The current $I$ through battery, $I=$ $\frac{8}{2}=4\text{A}$.

From the second diagram, the part $ED\text{A}\mathrm{C}D$ has a resistance of $4\Omega$, the total current will divide equally between $ED\text{A}\mathrm{C}D\&E\text{A}$. So consider $I_1=2\text{A}$ flowing through $ED$. Note that $I_1$ will be the same current flowing in the first diagram through $ED$.

From first diagram, when the resistances are in parallel the current is divided in the inverse ratio. Thus,
$I^{\prime }$ through $CD=\frac{3}{9}\times 2=\frac{2}{3}\text{A}$
I ' ' through $R_2=\frac{1}{2}\left(\frac{2}{3}\right)=\frac{1}{3}\text{A}$