Question

Q. A current of 0.1 A enters a Wheatstone bridge from a source of electricity. The resistance of each of the first three arms of the bridge is 10 Ω and the resistance of its fourth arm is 20 Ω. If the resistance of the galvanometer be 100Ω what will be the galvanometer current ?
​

Answer 1

Dear student,

take the two loop of the bridge at apply kirchoff loop law.
$$\begin{gathered} distributethecurrent. \\ currentthroughthefirst10ohmisi. \\ from100ohmisi-i\text{'} \\ forloop1 \\ 10i+100\left(i-i\text{'}\right)-10\left(0.1-i\right)=0 \\ 120i-100i\text{'}=1.....1 \\ forloop2 \\ 10i\text{'}-20\left(0.1-i\text{'}\right)-100\left(i-i\text{'}\right)=0 \\ 130i\text{'}-100i=2....2 \\ 600i-500i\text{'}=5 \\ 780i\text{'}-600i=12 \\ 280i\text{'}=17 \\ i\text{'}=17/280A=0.0607A \\ 120i-100\left(17/280\right)=1 \\ i=0.059A \\ currentthroughthegalvanometer=0.0607-0.059=0.001714A\approx 1.8mA \\ Regards \end{gathered}$$