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 \[I=\frac{{{E}_{1}}}{R}\Rightarrow P\frac{{{E}_{1}}{{E}_{2}}}{R}\cos \theta =\frac{E-E_{1}^{2}-E_{2}^{2}}{2R}\] |
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