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www.GOALias.blogspot.comPhysics1= = 4Ω62 × 3.14 × 50 × 796 × 10 −Therefore,Z = R + ( X − X ) = 3 + (8−4)2 2 2 2L C= 5 Ω(b) Phase difference, φ = tan –1 X − X C LREXAMPLE 7.8−1 ⎛4 − 8⎞= tan ⎜ = − 53.1°⎝⎟3 ⎠Since φ is negative, the current in the circuit lags the voltageacross the source.(c) The power dissipated in the circuit isP =Now,2I Ri m1 ⎛283⎞I = = ⎜ = 40A2 2 ⎝⎟5 ⎠Therefore, P = (40 A) 2 × 3 Ω = 4800 W(d) Power factor = cosφ = cos53.1° = 0.6Example 7.9 Suppose the frequency of the source in the previousexample can be varied. (a) What is the frequency of the source atwhich resonance occurs? (b) Calculate the impedance, the current,and the power dissipated at the resonant condition.Solution(a) The frequency at which the resonance occurs is1 1ω0= =LC 25.48 × 10 × 796 × 10= 222.1rad/s−3 −6ω0 221.1νr= = Hz = 35.4Hz2π 2 × 3.14(b) The impedance Z at resonant condition is equal to the resistance:Z= R = 3 Ω254EXAMPLE 7.9The rms current at resonance isV V ⎛283⎞1= = = 66.7AZ R⎜ =⎝⎟2 ⎠ 3The power dissipated at resonance isP I R2 2= × = × =(66.7) 3 13.35 kWYou can see that in the present case, power dissipatedat resonance is more than the power dissipated in Example 7.8.