question_answer

For the circuit shown in the figure the rms value of voltages across R and coil are \[{{E}_{1}}\]and \[{{E}_{2}}\]respectively. The power (thermal) developed

Across the coil is:

A) \[\frac{E-E_{1}^{2}}{2R}\]                  

B) \[\frac{E-E_{1}^{2}-E_{2}^{2}}{2R}\]

C) \[\frac{{{E}^{2}}}{2R}\]                                 

D) \[\frac{{{\left( E-{{E}_{1}} \right)}^{2}}}{2R}\]

Correct Answer: B

Solution :

draw the phasor diagram, \[{{\operatorname{E}}^{2}}={{E}_{1}}^{2}+{{E}_{2}}^{2}+2{{E}_{1}}{{E}_{2}}\cos \theta \]

Thermal power developed in coil is \[P={{E}_{2}}\cos \theta \times I\]and \[I=\frac{{{E}_{1}}}{R}\Rightarrow P\frac{{{E}_{1}}{{E}_{2}}}{R}\cos \theta =\frac{E-E_{1}^{2}-E_{2}^{2}}{2R}\]