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in Physics by (87.2k points)
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For the circuit shown in the figure the rms value of voltage across `R` and coil are `E_(1)` and `E_(2)` respectively.
The power (thermal) developed across the coil is
image
A. `(E - E_(1)^(2))/(2 R)`
B. `(E - E_(1)^(2) - E_(2)^(2))/(2 R)`
C. `(E^(2))/(2 R)`
D. `((E - E_(1))^(2))/(2R)`

1 Answer

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Best answer
Correct Answer - 2
Draw the phasor diagram
`E^(2) = E_(1)^(2) + E_(2)^(2) + 2E_(1) E_(2) cos theta`.
Thermal power developed in coil is
`P = E_(2) cos theta xx I` and
`I = (E_(1))/(R ) implies P = (E_(1) E_(2))/(R ) cos theta = (E^(2) - E_(1)^(2) - E_(2)^(2))/(2R)`
image

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