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(B) The circuit is a Wheatstone bridge with resistances $P = 10 \, \Omega$,$Q = 100 \, \Omega$,$R = 2 \, \Omega$,and $S = 20 \, \Omega$. The galvanome

Vedclass · Accepted Answer

$i, i_2, i_1, i_g$

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(B) The circuit is a Wheatstone bridge with resistances $P = 10 \, \Omega$,$Q = 100 \, \Omega$,$R = 2 \, \Omega$,and $S = 20 \, \Omega$. The galvanometer resistance is $G = 20 \, \Omega$.$1$. The total current $i$ is the sum of currents in the two branches: $i = i_1 + i_2$. Thus,$i$ is the largest current.$2$. Applying Kirchhoff's laws to the nodes,we find the potential distribution. The branch with lower resistance carries more current. Comparing the branches:- Branch $1$ (top): $10 \, \Omega$ and $100 \, \Omega$ in series.- Branch $2$ (bottom): $2 \, \Omega$ and $20 \, \Omega$ in series.$3$. Since the potential difference across both branches is the same $(2 \, V)$,and the total resistance of the bottom branch $(22 \, \Omega)$ is much smaller than the top branch $(110 \, \Omega)$,the current $i_2$ is greater than $i_1$.$4$. The galvanometer current $i_g$ flows from the higher potential point to the lower potential point. Given the bridge is unbalanced,calculation shows $i_g$ is the smallest among these.$5$. Therefore,the decreasing order is $i > i_2 > i_1 > i_g$.

In the circuit shown below,the resistance of the galvanometer is $20 \, \Omega$. In which of the following alternatives are the currents arranged strictly in decreasing order?

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