(a) Writing two loop equations (b) Calculation of currents through $40 \Omega$ and $20 \Omega$ resistors In loop $A B C D A$$+80-20 I_2+40 I_1\;=0$$4\;=I_2-2 I_1$In loop FCDEF$-40 I_1-10 (I_1+I_2) +40\;=0$$-50 I_1-10 I_2+40\;=0$$5 I_1+I_2\;=4$Solving these two equations$\;I_1=0 A$$\;I_2=4 A$ OR End error, overcoming Formula for metre bridge Calculation of value of $S$ The end error, in a metre bridge, is the error arising due to (i) Ends of the wire not coinciding with the $0 \text{cm} / 100 \text{cm}$ marks on the metre scale. (ii) Presence of contact resistance at the joints of the meter bridge wire with the metallic strips. It can be reduced/overcome by finding balance length with two interchanged positions of $R$ and $S$ and taking the average value of ' $S$ ' from two readings. For a metre bridge $\;\frac{R}{S}=\;\frac{l}{100-l}$ For the two given conditions $\;\frac{5}{S}=\;\frac{l_1}{100-l_1}$ $\;\frac{5}{\frac{S}{2}}=\;\frac{1.5 l_1}{100-1.5 l_1}$Dividing the two equations$2\;=\;\frac{1.5 l_1}{100-1.5 l_1} \times \;\frac{100-l_1}{l_1}$$200-3 l_1\;=150-1.5 l_1$$l_1\;=\;\frac{100}{3} \text{cm}$ Putting the value of $l_1$ in any one of the two given conditions.$S=10 \Omega$