In the circuit shown,wire 1 is of negligible | vedclass

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(D) Consider the two loops separately. In the first loop,the current $i_1$ flows through the resistor $R_1$ and the cell $\varepsilon_1$. By Kirchhoff

Vedclass · Accepted Answer

no current will flow through wire $1$

Vedclass · Answer

(D) Consider the two loops separately. In the first loop,the current $i_1$ flows through the resistor $R_1$ and the cell $\varepsilon_1$. By Kirchhoff's voltage law,$i_1 = \frac{\varepsilon_1}{R_1}$.Similarly,in the second loop,the current $i_2$ flows through the resistor $R_2$ and the cell $\varepsilon_2$. By Kirchhoff's voltage law,$i_2 = \frac{\varepsilon_2}{R_2}$.Since each loop is a closed circuit independent of the other,the charge conservation principle implies that the current entering any junction must equal the current leaving it. For the junction connecting the two loops to wire $1$,if we apply Kirchhoff's current law,the sum of currents must be zero. Since the loops are isolated from each other regarding current flow,no current can flow through the connecting wire $1$ to maintain the continuity of the individual loops. Thus,the current through wire $1$ is zero.

In the circuit shown,wire $1$ is of negligible resistance. Then,

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