Question

Which arrangement (series or parallel) takes more current from the battery? Give answer with reasons

Answer 1

This can be be easily analyzed by the series and parallel combination law of resistors.
$\text{In series,}\\ R=\sum_{r=1}^{n}R_i\\ \Rightarrow R \ge R_j\\ \text{where Rj is one of the resistors}\\ \Rightarrow \frac{1}{R} \le \frac{1}{R_j}\\ \text{Multiplying both sides with V,}\\ \Rightarrow \frac{V}{R} \le \frac{V}{R_j}\\ \Rightarrow I \le I_j\\ \text{where Ij is the current flowing when Rj is individually connected to the battery.}\\ \text{In parallel,}\\ \frac{1}{R}=\sum_{r=1}^{n}\frac{1}{R_i}\\ \Rightarrow \frac{1}{R} \ge \frac{1}{R_j}\\ \text{where Rj is one of the resistors}\\ \text{Multiplying both sides with V,}\\ \Rightarrow \frac{V}{R} \ge \frac{V}{R_j}\\ \Rightarrow I \ge I_j\\ \text{where Ij is the current flowing when Rj is individually connected to the battery.}$

Answer 2

In series circuit, the effective resistance is equal to sum of the resistances of individual components. So total resistance will be on higher side. In parallel circuit, reciprocal of effective resistance is equal to sum of reciprocals of individual resitances. So effective resistance is less.

So, clearly parallel combination takes more current from the battery.

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