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The coil of an a.c. generator has 100 turns, each of cross sectional area `2 m^(2)`. It is rotating at a constant angular speed of 30 radians`//`s, in a uniform magnetic field of `2 xx 10^(-2)`T. What is the maximum power dissipated in the circuit, if the resistance of the circuit including that of the coil is `600 Omega`?A. 6 WB. 9 WC. 12 WD. 24 W |
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Answer» Correct Answer - C `N = 1000,A = 2 m^(2), omega = 30 rad//s, T = 2 xx 10^(-2)T`, `R = 600 Omega` Maximum power dissipated in the circuit, `P_(max) = E_(rms) xx I_(rms) = (E_0)/(sqrt2) xx (I_0)/(sqrt2) = (E_(0)I_(0))/(2)` But `I_(0) = (E_0)/R` `:. P_(max) = (E_(0) cdot E_(0))/(R xx 2) = (E_(0)^(2))/(2R) but E_(0) = NAB omega` `:. P_(max) = ((NABomega)^(2))/(2R)` `=((100 xx 2 xx 2 xx 10^(-2) xx 30)^(2))/(2 xx 600)` `:. P_(max) = (120 xx 120)/(2 xx 600) = 12 W`. |
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