Given circuit below



Resistance between $B$ and $C$ on the right-hand side of the circuit

$=\frac{10\times 10}{10+10}=\frac{100}{20}=5\mathrm{\Omega }$

because $1\mathrm{\Omega }+4\mathrm{\Omega }+5\mathrm{\Omega }=10\mathrm{ }\mathrm{\Omega }$ are in parallel with $10 \mathrm{\Omega }$ .

Resistance between $A$ and $D$ on the right-hand side of the circuit.

$=\frac{10\times 10}{10+10}=\frac{100}{20}=5\mathrm{\Omega }$

because $5\mathrm{\Omega }$ (equivalent resistance of $1\mathrm{\Omega },\mathrm{ }4\mathrm{\Omega },\mathrm{ }5\mathrm{\Omega }$ and $10 \mathrm{\Omega }$ ) is in series with $3\mathrm{\Omega }$ and $2\mathrm{\Omega }$ .

Hence, resistance between $A$ and $D$ will be $5\mathrm{\Omega }$

Equivalent resistance of the circuit,

$R=5+2+5=12\mathrm{\Omega }$

Current drawn from the battery,

$I=\frac{12V}{12\mathrm{\Omega }}=1\text{A}$

At the junction $\text{A}$ , the current of $1 \text{A}$ is divided equally between the $10\mathrm{\Omega }$ resistance and the remaining circuit of the $10\mathrm{\Omega }$ resistance.

At the junction $B$ , the current of $0.5 \text{A}$ is divided equally between the $10\mathrm{\Omega }$ resistor and remaining circuit of resistance $10\mathrm{\Omega }$ .

$\therefore$ Current through the $4\mathrm{\Omega }$ resistor .