Physics - Current Electricity Question with Solution | TestHub

In the circuit, the branch EGHF have three resistance of $10\Omega ,20\Omega$ and $10\Omega$, respectively which are connected in series combination. So, their equivalent resistance is given by ${\mathrm{R}}_1=10+20+10=40\Omega$ This $R_1$ resistance is parallel with $40\Omega$ resistance which is connected in the branch EF. So, their equivalent resistance ${\mathrm{R}}_2=\frac{40\times 40}{40+40}=20\Omega$ Now, circuit becomesNow, t in the branch CEFD $10\Omega ,20\Omega$ and $10\Omega$ resistance are connect in series combination their equivalent resistance is given by ${\mathrm{R}}_3=10\Omega +20\Omega +40\Omega =40\Omega$ This $R_3$ is parallel with $40\Omega$ resistance which is present in branch CD. Their equivalent resistance, ${\mathrm{R}}_4=\frac{40\times 40}{40+40}=20\Omega$ Now, circuit becomesThe net resistance between A and B , ${\mathrm{R}}_{\text{net }}=10+20+10=40\Omega$