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Sinusoidal Steady State-MCQs 1. The value of current through the 1 Farad capacitor of figure is 0.5F 1 1 1F 2 2 sin 100 t AC 1H 1 1 0.5F (a) zero (b) one (c) two (d) three [GATE 1987: 2 Marks] Sol. The given circuit is a bridge circuit and can be redrawn as 2 sin 100 t 2+s AC a 1F b 1+2/s The product of the opposite arms are equal ๐ฝ๐ = ๐ฝ๐ ๐ฝ๐ = ๐ฝ๐ = ๐ ๐ฌ๐ข๐ง ๐๐๐๐ × ๐ = ๐ฌ๐ข๐ง ๐๐๐๐ ๐ ๐ ๐ฌ๐ข๐ง ๐๐๐๐ ๐ × (๐ + ) = ๐๐๐๐๐๐๐ ๐ ๐(๐ + ๐ ) ๐ The current through 1F capacitor is zero Option (a) 2. The half โ power bandwidth of the resonant circuit of figure can be increased by: R1 R2 C L (a) increasing R1 (b) decreasing R1 Sol. (c) increasing R2 (d) decreasing R2 [GATE 1989: 2 Marks] ๐ Bandwidth of the circuit โ ๐ธ For increasing bandwidth, Q is to be decreased ๐๐ณ ๐ ๐ธ= = ๐น๐ ๐๐ช๐น๐ If R1 โ 0, R2โโ, the circuit has only L & C elements and has high selectivity. ๐น๐ โ, ๐น๐ โ, ๐ธ โ ๐ฉ๐พ โ Option (a) & (d) 3. The resonant frequency of the series circuit shown in figure is M=1H 2H (a) (b) 1 ๐ป 4๐โ3 ๐ 1 ๐ป 4๐ ๐ 2H 2F (c) (d) 1 2๐โ10 1 4๐โ2 ๐ป๐ ๐ป๐ [GATE 1990: 2 Marks] Sol. The coils are connected in a series opposing way. As the current is entering the dot of coil L1 and leaving the dot of coil L2 ๐ณ๐๐ = ๐ณ๐ + ๐ณ๐ โ ๐๐ด =๐+๐โ๐ = ๐๐ฏ At resonance XL = XC ๐ ๐ณ๐๐ = ๐= ๐ โ๐ณ๐๐ ๐ช = ๐ ๐๐ช ๐ โ๐ × ๐ ๐= = ๐ ๐๐๐ /๐๐๐ ๐ ๐ ๐ฏ๐ ๐๐ Option (b) 4. In the series circuit shown in figure, for series resonance, the value of the coupling coefficient k will be K 18 -j12 j2 (a) 0.25 (b) 0.5 j8 (c) 0.999 (d) 1.0 [GATE 1993: 1 Mark] Sol. The coils are connected in series additive manner as the current is entering the dot in both coils ๐ณ๐๐ = ๐ณ๐ + ๐ณ๐ + ๐๐ด ๐ด = ๐ฒโ๐ณ๐ ๐ณ๐ = ๐ฒโ๐๐. ๐๐ = ๐ฒ๐๐ At resonance |๐ฟ๐ณ | = |๐ฟ๐ช | ๐ฟ๐ณ = ๐ฟ๐ณ๐ + ๐ฟ๐ณ๐ + ๐ฟ๐ด = ๐๐ + ๐๐ + ๐๐ฒ๐๐ ๐๐๐ + ๐๐ฒ๐๐ At resonance ๐ฟ๐ณ = ๐ฟ๐ช ๐๐๐ + ๐๐ฒ๐๐ = ๐๐๐ ๐๐ฒ๐. ๐ = ๐. ๐ ๐๐ฒ. ๐ = ๐. ๐ ๐ฒ= ๐ ๐ = = ๐. ๐๐ ๐ ๐ Option (a) 5. In figure, A1, A2, and A3 are ideal ammeters. If A1 reads 5A, A2 reads 12A, then A3 should read R A1 A3 C A2 AC 100 sin ฯt (a) 7 A (b) 12 A (c) 13 A (d) 17 A [GATE 1993: 2 Marks] Sol. Since the source is a.c. ๐ฐ๐(๐๐๐) = โ๐ฐ๐๐ (๐๐๐) + ๐ฐ๐๐ (๐๐๐) = โ๐๐ + ๐๐๐ = โ๐๐๐ = ๐๐๐จ Option (c) 6. A series LCR circuit consisting of ๐ = 10 โฆ, |๐๐ฟ | = 20 โฆ ๐๐๐ |๐๐ถ | = 20 โฆ is connected across an a.c. supply of 200 V rms. The rms voltage across the capacitor is (a) 200โ โ 900 ๐ (c) 400โ + 900 ๐ (b) 200โ + 900 ๐ (d) 400โ โ 900 ๐ [GATE 1994: 1 Mark] Soln. For a series LCR circuit ๐ = ๐น + ๐๐ฟ๐ณ โ ๐๐ฟ๐ช ๐๐๐๐๐ ๐ฟ๐ณ = ๐ฟ๐ช , ๐ฝ ๐ฐ=๐น= ๐๐๐ ๐๐ ๐=๐น = ๐๐๐จ The voltage across the capacitor= ๐ฐ. (โ๐๐ฟ๐ช ) = ๐๐. (โ๐๐๐) = โ๐๐๐๐ = ๐๐๐โ โ ๐๐๐ ๐ฝ Option (d) 7. A DC voltage source is connected across a series R-L-C circuit. Under steady state conditions, the applied DC voltage drops entirely across the (a) R only (b) L only Soln. (c) C only (d) R and L combination [GATE 1995: 1 Mark] Under steady state condition, inductor behaves as a short circuit and capacitor as an open circuit. The applied voltage drops entirely across the capacitor R Option (C) L C 8. Consider a DC voltage source connected to a series R-C circuit. When the steady state reaches, the ratio of the energy stored in the capacitor to the total energy supplied by the voltage source, is equal to (a) 0.362 (c) 0.632 (b) 0.500 (d) 1.000 [GATE 1995: 1 Mark] Soln. The total energy supplied by the source ๐พ๐บ = ๐ฝ๐ฐ. ๐ ๐ = ๐ฝ โซ ๐ฐ๐ ๐ = ๐ฝ๐ ๐ = ๐ฝ๐ช๐ฝ = ๐ช๐ฝ๐ Voltage across the uncharged capacitor ๐ฝ๐ช (๐) = ๐ฝ๐บ (๐ โ ๐ โ๐โ ๐น๐ช ) Power in the capacitor = ๐ฝ๐บ (๐ โ ๐ = ๐ฝ๐๐บ ๐น (๐ โ๐โ ๐น๐ช โ๐โ ๐ ๐น๐ช ) ๐ช ๐ฝ (๐ ๐ ๐ ๐บ โ๐ โ๐ โ๐โ ๐น๐ช ) โ๐๐โ ๐น๐ช ) Energy stored in the capacitor at steady state โ โ ๐ฝ๐๐บ ๐ฝ๐๐บ โ๐โ โ๐๐ ๐น๐ช โซ๐ ๐ ๐ โ โซ ๐ โ๐น๐ช ๐ ๐ ๐น ๐น ๐ = ๐ฝ๐๐บ ๐น [๐น๐ช โ ๐ ๐น๐ช ๐ ] ๐ = ๐ ๐ช๐ฝ๐๐บ ๐๐๐๐๐๐ ๐๐๐๐๐๐ ๐๐ ๐๐๐๐๐๐๐๐๐๐๐๐ = ๐. ๐ ๐๐๐๐๐๐ ๐๐๐๐๐๐๐๐ ๐๐ ๐๐๐ ๐๐๐๐๐๐ Option (b) 9. The current, i(t), through a 10-โฆ resistor in series with an inductance, is given by i(t) = 3 + 4 sin (100 t + 450) + 4 sin (300 t + 600) amperes. The RMS value of the current and the power dissipated in the circuit are: (c) 5 A, 250 W, respectively (d) 11 A, 1210 W respectively (a) โ41A, 410 W, respectively (b) โ35 A, 350 W, respectively [GATE 1995: 1 Mark] Soln. ๐(๐) = ๐ + ๐ ๐ฌ๐ข๐ง(๐๐๐๐ + ๐๐๐ ) + ๐ ๐ฌ๐ข๐ง(๐๐๐๐ + ๐๐๐ ) ๐ ๐ ๐ ๐ ๐ฐ๐๐๐ = โ๐๐ + ( ) + ( ) = ๐ ๐๐๐๐ โ๐ โ๐ Power dissipated = ๐ฐ๐๐๐๐ ๐น = (๐)๐ × ๐๐ = ๐๐๐ ๐๐๐๐๐ Option (c) 10. In the circuit of Figure assume that the diodes are ideal and the meter is an average indicating ammeter. The ammeter will read A + D2 4 sin (ฯt) Volts 10 k 10 k (a) 0.4 โ2 mA (b) 0.4 mA (c) (d) 0.8 ๐ 0.4 ๐ ๐๐ด ๐๐ด [GATE 1996: 1 Mark] Soln. For positive half cycle of input D1 is conducting and D2 is off The current through ammeter is the average value of positive half sine wave i.e. ๐ฐ๐๐๐ = ๐ฝ๐ ๐ ๐ ๐. ๐ × = = ๐๐จ ๐ ๐ ๐น ๐ × ๐๐ × ๐๐ ๐ Option (d) 11. A series RLC circuit has a resonance frequency of 1 KHz and a quality factor Q = 100. If each of R, L and C is doubled from its original value, the new Q of the circuit is (a) 25 (b) 50 (c) 100 (d) 200 [GATE 2003: 1 Mark] Soln. ๐ธ= = ๐๐ณ ๐ณ = ๐๐ ๐๐ ๐น ๐น ๐๐ ๐ณ ๐ ๐ณ = โ ๐๐ โ๐ณ๐ช ๐น ๐น ๐ช When R, L and C are doubled ๐ธโฒ = ๐ธ ๐ Option (b) 12. An input voltage ๐ฃ(๐ก) = 10โ2 cos(๐ก + 100 ) + 10โ5 cos(2๐ก + 100 ) V is applied to a series combinationof R = 1 โฆ and an inductance L = 1H. The resulting steady state current i(t) in ampere is (a) 10 cos(๐ก + 550 ) + 10 cos(2๐ก + 100 + ๐ก๐๐โ1 2) 3 (b) 10 cos(๐ก + 550 ) + 10โ2 cos(2๐ก + 55) (c) 10 cos(๐ก โ 350 ) + 10 cos(2๐ก + 100 โ ๐ก๐๐โ1 2) 3 (d) 10 cos(๐ก โ 350 ) + 10โ2 cos(2๐ก โ 35) [GATE 2003: 2 Marks] Soln. ๐ฝ(๐) = ๐๐โ๐ ๐๐จ๐ฌ(๐ + ๐๐๐ ) + ๐๐โ๐ ๐๐จ๐ฌ(๐๐ + ๐๐๐ ) ๐ฝ(๐) = ๐๐โ๐ ๐๐จ๐ฌ(๐ + ๐๐๐ ) + ๐๐โ๐ ๐๐จ๐ฌ(๐๐ + ๐๐๐ ) ๐๐ = ๐, ๐๐ = ๐ Steady state current ๐(๐) = ๐๐ (๐) + ๐๐ (๐) ๐๐โ๐ ๐๐จ๐ฌ(๐ + ๐๐๐ ) ๐๐โ๐ ๐๐จ๐ฌ(๐๐ + ๐๐๐ ) = + ๐น + ๐๐๐ ๐ณ ๐น + ๐๐๐ ๐ณ = = ๐๐โ๐ ๐๐จ๐ฌ(๐ + ๐๐๐ ) ๐๐โ๐ ๐๐จ๐ฌ(๐๐ + ๐๐๐ ) + ๐+๐ ๐ + ๐๐ ๐๐โ๐ ๐๐จ๐ฌ(๐ + ๐๐๐ ) โ๐โ ๐๐๐ + ๐๐โ๐ ๐๐จ๐ฌ(๐๐ + ๐๐๐ ) โ๐โ ๐ญ๐๐งโ๐ ๐ = ๐๐ ๐๐จ๐ฌ(๐ โ ๐๐๐ ) + ๐๐ ๐๐จ๐ฌ(๐๐ + ๐๐๐ โ ๐ญ๐๐งโ๐ ๐) Option (C) 1 1 13. The circuit shown in the figure, with ๐ = 3 โฆ, ๐ฟ = 4 ๐ป, ๐ถ = 3 F has input voltage V(t) = Sin2t. The resulting current i(t) is i(t) V(t) (a) 5 sin(2๐ก + 53. 10 ) (b) 5 sin(2๐ก โ 53. 10 ) R L C (c) 25 sin(2๐ก + 53. 10 ) (d) 25 sin(2๐ก โ 53. 10 ) [GATE 2003: 1 Mark] Soln. Admittance ๐= ๐ ๐ + + ๐๐๐ช ๐น ๐๐๐ณ =๐+ ๐ + ๐๐ × ๐ ๐๐ = ๐ โ ๐๐ + ๐๐ = ๐ + ๐๐ ๐(๐) = ๐ฝ(๐). ๐ = (๐ + ๐๐) ๐ฌ๐ข๐ง ๐๐ = ๐ ๐ฌ๐ข๐ง ๐๐ โ ๐ญ๐๐งโ๐ ( ๐/๐) = ๐ ๐ฌ๐ข๐ง(๐๐ + ๐๐. ๐๐ ) Option (a) 14. For the circuit shown in the figure, the time constant RC = 1 ms. The input voltage is ๐๐ (๐ก) = โ2 sin 103 ๐ก. The output voltage V0 (t) is equal to R vi(t) v0(t) (a) sin(103 ๐ก โ 450 ) (b) sin(103 ๐ก + 450 ) Soln. (c) sin(103 ๐ก โ 530 ) (d) sin(103 ๐ก + 530 ) [GATE 2004: 1 Mark] ๐ฝ๐ (๐) = โ๐ ๐ฌ๐ข๐ง ๐๐๐ ๐ ๐ = ๐๐๐ ๐๐๐ ๐๐๐๐/๐๐๐ ๐น๐ช = ๐๐ ๐๐๐ ๐ฝ๐ (๐) = ๐ฝ๐ (๐) ๐ (๐น + ๐๐๐) × ๐ ๐๐๐ ๐ฝ๐ (๐) โ๐ ๐ฌ๐ข๐ง ๐๐๐ ๐ = = ๐๐๐น๐ช + ๐ ๐๐๐๐ × ๐๐โ๐ + ๐ = โ๐ ๐ฌ๐ข๐ง ๐๐๐ ๐ ๐+๐ = ๐ฌ๐ข๐ง(๐๐๐ ๐ โ ๐๐๐ ) Option (a) 15. Consider the following statements S1 and S2 S1: At the resonant frequency the impedance of a series R-L-C circuit is zero S2: In a parallel G-L-C circuit, increasing the conductance G results in increase in its Q factor. Which one of the following is correct? (a) S1 is FALSE and S2 is TRUE (b) Both S1 and S2 are TRUE Soln. (c) S1 is TRUE and S2 is FALSE (d) Both S1 and S2 are FALSE [GATE 2004: 2 Marks] The impedance of a series RLC circuit at resonant frequency is ๐ ๐ = ๐น + ๐ (๐๐ณ โ ๐.๐ช) ๐ ๐๐ณ = ๐.๐ช At resonance So, Z = R purely resistive In a parallel RLC circuit, ๐ 3dB bandwidth๐๐ โ ๐๐ = ๐น๐ช ๐ธ= ๐๐ ๐๐ = = ๐๐น ๐น๐ช ๐๐ โ ๐๐ ๐โ ๐น๐ช ๐๐ is the resonant frequency. As conductance ๐ฎ โ ๐น โ ๐ธ โ (As conductance G increases R reduces so Q reduces) Option (d) ZL = (7+4j) 20 00V 16. An AC source of RMS voltage 20 V with internal impedance ๐๐ = (1 + 2๐)โฆ feeds a load of impedance ๐๐ฟ = (7 + 4๐)โฆ in the figure below. The reactive power consumed by the load is ZS = (1+2j) DC (a) 8 VAR (b) 16 VAR (c) 28 VAR (d) 32 VAR [GATE 2009: 2 Marks] Soln. The current I in the given circuit is ๐ฐ= = = ๐๐ ๐๐ = ๐ + ๐๐ ๐ + ๐๐ ๐๐ โ๐๐ + ๐๐ โ ๐ญ๐๐งโ๐ ๐/๐ ๐๐ = ๐โ โ ๐๐. ๐๐๐ ๐โ ๐๐. ๐๐ Reactive power = ๐ฐ๐ ๐ฟ๐ณ = (๐)๐ × ๐ = 16 VAR Option (b)