A) \[1:4\]
B) \[4:1\]
C) \[2:1\]
D) \[2:1\]
Correct Answer: A
Solution :
(i) In series : total resistance\[=R+R=2R\] Power consumed \[{{P}_{1}}=\frac{{{V}^{2}}}{2R}\] ... (i) (ii) In parallel: potential difference across each resistance will be\[V\] So, power consumed in each resistance is \[P=\frac{{{V}^{2}}}{R}\] So total power consumed in two resistance is \[{{P}_{2}}=\frac{{{V}^{2}}}{R}+\frac{{{V}^{2}}}{R}=\frac{2{{V}^{2}}}{R}\] ? (ii) Hence, ratio in two cases \[=\frac{{{P}_{1}}}{{{P}_{2}}}\] \[=\frac{{{V}^{2}}\times R}{2R\times 2{{V}^{2}}}\] \[=\frac{1}{4}\]
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