Find $R_{\text {net }}$ between $A$ and $B$.

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 becomes Now, 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 becomes The net resistance between A and B , $\mathrm{R}_{\text {net }}=10+20+10=40 \Omega$