A) 1 : 4
B) 4 : 1
C) 2 : 1
D) 1 : 2
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, required ratio\[=\frac{{{P}_{1}}}{{{P}_{2}}}\] \[=\frac{{{V}^{2}}\times R}{2R\times 2{{V}^{2}}}=\frac{1}{4}\]
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