A) \[{{P}_{1}}+{{P}_{2}}watt\]
B) \[\sqrt{{{P}_{1}}{{P}_{2}}}\]watt
C) \[\frac{{{P}_{1}}{{P}_{2}}}{{{P}_{1}}+{{P}_{2}}}watt\]
D) \[\frac{{{P}_{1}}+{{P}_{2}}}{{{P}_{1}}{{P}_{2}}}watt\]
Correct Answer: A
Solution :
If resistances of bulbs are \[{{R}_{1}}\] and \[{{R}_{2}}\] respectively then in parallel \[\frac{1}{{{R}_{P}}}=\frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}}\]Þ \[\frac{1}{\left( \frac{{{V}^{2}}}{{{P}_{p}}} \right)}=\frac{1}{\left( \frac{{{V}^{2}}}{{{P}_{1}}} \right)}+\frac{1}{\left( \frac{{{V}^{2}}}{{{P}_{2}}} \right)}\] Þ \[{{P}_{P}}={{P}_{1}}+{{P}_{2}}\]
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