A) the current through the circuit is\[\frac{n}{k}\].
B) the potential difference between the terminals of the 1st battery is zero.
C) the current through the circuit is \[\frac{{{n}^{2}}}{k}\].
D) the potential difference between the terminals of the / battery is\[\frac{\varepsilon }{k}\].
Correct Answer: B
Solution :
[b]: Suppose the current is I in the indicated direction. Applying Kirchoff?s loop law, \[{{\varepsilon }_{1}}-I{{r}_{1}}+{{\varepsilon }_{2}}-I{{r}_{2}}+{{\varepsilon }_{3}}-I{{r}_{3}}+...+{{\varepsilon }_{n}}-I{{r}_{n}}=0\] or,\[I=\frac{{{\varepsilon }_{1}}+{{\varepsilon }_{2}}+{{\varepsilon }_{3}}+...+{{\varepsilon }_{n}}}{{{r}_{1}}+{{r}_{2}}+{{r}_{3}}+...{{r}_{n}}}\] \[=\frac{{{\varepsilon }_{1}}+{{\varepsilon }_{2}}+{{\varepsilon }_{3}}+...+{{\varepsilon }_{n}}}{k({{\varepsilon }_{1}}+{{\varepsilon }_{2}}+{{\varepsilon }_{3}}+...{{\varepsilon }_{n}}}=\frac{1}{k}\] The potential difference between the terminals of the 1th battery is\[{{\varepsilon }_{1}}-I{{r}_{i}}={{\varepsilon }_{i}}-\left( \frac{1}{k} \right)(k{{\varepsilon }_{i}})=0\]
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