Alternating Current-08- (Solutions)

Alternating Current (Solutions) LEVEL I OBJECTIVE 1. 2. 3. at and and graph will be rectangular hyperbola. 4 V = V0 (sin t) Impedance 5 Impedance phase difference (between ) 6. Current I in the circuit is given by Final current is at 7. …(i) …(ii) 8. Frequency will remain unchanged 9. Average power 10. Z = Impedance V Objective Unsolved level – II 1. …(i) I = 2. Current Reading of ammeter =20 mA 3. Power factor cos  = But at resonance 4. Phase difference at Resonance 5. As squaring I will be maximum of 6. (Mean square current ) = = 7. 8. 9. The root mean square voltage is 10. Let R is resistance of inductor Now for AC supply solving = Subjective Unsolved CBSE level - 1 1. rad s-1 rad s-1. 2. when frequency of supply is equal to natural frequency phase difference 3.(a) (b) (c) Vrms (capacitor) = and 4.(a) = (b) (c) here (i) energy will be completely electrical if i.e. (ii) Completely magnetic If where (d) As per question n ; 1, 2, 3 (e) the whole initial energy will damp. 5.(a) (b) 6.(a) (b) Similar as that of Q. No. 5 (b). 7. Total emf applied across the lamp + pd V2 across choke coil or current will be same in the circuit, hence the pd across the choke of self inductance, L is or 8. C = 60 F = 60× 10-6 F (a) = = (b) (c) zero (d) zero (e) zero 9.(a) current I will be maximum when and (b) Power P = VrmsIrms cos  and it will be maximum if i.e. phase difference is zero and (c) As per question solving and (d) Q (factor) = 10. Impedance current . Power component of current = Wattles component of current = Subjective Unsolved Level – II 1. Putting respective valves and so on. 2.(a) (b) 3.(a) Current will be maximum for resonating circuit. i.e. for F (Frequency) = (b) 4. There is no change in flux ( constant DC supply) in secondary coil induced voltage and current will be zero 5.(a) (b) (source) (c) 6. Peak current = instantaneous current = 7. 8. Now when AC source is applied 9. Effective (rms) current Power factor 10. will be zero after inserting a capacitor in a series with circuit. With value of C = Subjective Unsolved Level - III 1. 2. Let resistance of inductor be R for Irms to be maximum and (maximum) = Now for DC circuit 3. 4. where as the energy will only be dissipated in the circuit resistance hence average rate of consumption of energy = 5. Since AC voltmeter is of high impedance, it will draw zero current and reading will be same in each case and will be equal to 6. …(i) Now when they are connected in series. for resonance to accrue phase difference i.e. …(ii) Putting eqn. (i) and (ii) or 7. Total resistance in the circuit = 50 + 2 = 52  Inductive reactance Capacitive reactance = 79.6  as hence the current lags behind the applied emf by an angle , given by or . Power factor cos  = Apparent power = and true power = 8.(i) Impedance (ii) 9. Current amplitude = with 10.(a) For D.C connection …(i) for AC connection …(ii) with and eq. (ii) =80 mH (b) Energy will dissipated in Resistor only watt

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