Solution - Career Point

8 QuestionsSolutionSet # 11Physics Challenging Problemswere Published in March Issue1.[C] The given circuit as an R-L-C series circuit whenfrequency of the source varies the impedance of theR-L-C series circuit varies and correspondingly thecurrent in the circuit get variedImpedance variation and current variation are shownin figure.R = Zminimumf 1 f=fr f 2 fI =I0II 0f 1 f=fr f 2 f23[C] At frequency f 1 currentI 1 V 1 200 20± 0= . = . = = 102 2 R 2 10 2Watt less currentI µ = Isinφ1I µ = 10 2. = 10 Amp.2As φ = 45ºRbecause cos φ = R / Z = =R 22 Amp.V V 2004[A] At frequency current I = = = = 20 Amp.Zm R 10Potential difference across capacitor12∆f = f 2 -f 1∆f = f 2 -f 1V C = I C .X C = 1.X C = 20.X CAt frequency f 1 X C > X L Power factor – leadingnature of circuit is capacitanceAt frequency f 2 X L > X C Power factor – leadingnature of circuit is inductingAt frequency f 1 and frequency f 2 impedance Z=Because of the fact –V VAs I 0 = = ........(i)Zhm R⇒± =I 0 V=2 ZV / 2 V =2 Z2 . R⇒V V =R 2 Zf 1 < f < f 2and R 22.[A] ∆ f = f 2 − f1= B and width of R-L-C series circuit1= .R / L2πCharge on capacitor Q C = C.V C = C. 20X C1= C. (2v)ωC5.[C] Longest wavelengthλ6.[B] f20 20= =ω 2πf= (5π)= (5π)−1−120 1= =2π(50)5πCoulombcbv + vs350 × 0.8 × 5= =f 600v= fv − vmax =maxs350== 607Hz350 − 0.8 × 57.[A] 345.5/346.0 × 600 = 599 Hz8.[A]dydx= −1 dy d y 1& = +v dt2 2dx v2d2dty20.59mXtraEdge for IIT-JEE 16 APRIL 2010