question_answer

On a bulb is marked 60 volt and 10 watt and is to be run with an A.C. source of 100 volt. Then inductance of series coil inquired is:

A)  1.53 H  

B)                         2.15 H

C)         3.27 H                  

D)                                         3.89 H

Correct Answer: A

Solution :

                \[{{V}^{2}}={{V}_{R}}^{2}+B{{V}_{L}}^{2}\] \[V=\sqrt{{{V}^{2}}-{{V}_{R}}^{2}}=\sqrt{{{(100)}^{2}}-{{(60)}^{2}}}\]                 \[P=Vi\]                 \[10=60\times i=80volt\]                 \[i=\frac{1}{6}A\]                 \[{{X}_{L}}=\frac{V}{i}=\frac{80}{1/6}=480\Omega \]                 \[2\pi \,fL=480\]                 \[L=1.53H\]