A) 1000 W
B) 2000 W
C) 500 W
D) 4000 W
Correct Answer: B
Solution :
Let the resistance of the two heaters be denoted by \[{{R}_{1}}\] and \[{{R}_{2}}\]. Then, \[{{R}_{1}}=\frac{{{V}^{2}}}{{{P}_{2}}}\] \[{{R}_{2}}=\frac{{{V}^{2}}}{{{P}_{2}}}\] In parallel combination of resistances \[\frac{1}{{{R}_{p}}}=\frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}}\] \[\frac{{{P}_{p}}}{{{V}^{2}}}=\frac{{{P}_{1}}}{{{V}^{2}}}+\frac{{{P}_{2}}}{{{V}^{2}}}\] \[{{P}_{p}}={{P}_{1}}+{{P}_{2}}\] Given, \[{{P}_{1}}=1000\,\,W\] \[{{P}_{2}}=1000\,\,W\] \[\therefore \] \[{{P}_{p}}=1000+1000\] Hence, \[{{P}_{p}}=2000\,\,W\]
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