A) 1/4
B) 4
C) 3/2
D) 3
E) 1
Correct Answer: C
Solution :
In series, \[{{p}_{s}}=\frac{{{P}_{1}}\times {{P}_{2}}}{{{P}_{1}}+{{P}_{2}}}=12W\] In parallel, \[{{P}_{p}}={{P}_{1}}+{{P}_{2}}=50\,W\] \[\therefore \] \[{{P}_{1}}{{P}_{2}}=12\times 50=600\] Now, \[{{({{p}_{1}}-{{p}_{2}})}^{2}}={{({{p}_{1}}+{{p}_{2}})}^{2}}-4{{p}_{1}}{{p}_{2}}\] \[={{(50)}^{2}}-4\times 600\] \[=2500-2400=100\] \[\therefore \] \[{{P}_{1}}-{{P}_{2}}=10\] or \[{{P}_{1}}=30W,{{P}_{2}}=20\,W\] \[\therefore \] \[\frac{{{P}_{1}}}{{{P}_{2}}}=\frac{3}{2}\]You need to login to perform this action.
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