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$An alternating current is given as i = i_{1} \cos ωt - i_{2} \sin ωt . The rms current is given by$

$\frac{i_{1} + i_{2}}{\sqrt{2}}$

$\frac{|i_{1} + i_{2}|}{\sqrt{2}}$

$\sqrt{\frac{i_{1}^{2} + i_{2}^{2}}{2}}$

$\sqrt{\frac{I_{1}^{2} + i_{2}^{2}}{\sqrt{2}}}$

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Explanation

$i = i_{1} cosωt - i_{2} \cos (ωt - \frac{π}{2}) Here, angle between i_{1} and i_{2} = \frac{π}{2} i_{net} = \sqrt{{i_{1}}^{2} + {i_{2}}^{2} + 2 i_{1} i_{2} \cos \frac{π}{2}} = \sqrt{{i_{1}}^{2} + {i_{2}}^{2}} i_{rms} = \sqrt{\frac{{i_{1}}^{2} + {i_{2}}^{2}}{2}}$

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In a step-up transformer, the turn ratio is 1: 2. A Leclanche cell (e.m.f. 1.5V) is connected across the primary. The voltage developed in the secondary would be

3.0 v

0.75 v

1.5 v

Zero

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Explanation

If d.c. a source connected to the primary coil of the transformer then output is zero because of transformer work only A.C. not for D.C.

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A 220 V, 50 Hz ac source is connected to an inductance of 0.2 H and a resistance of 20 ohms in series. What is the current in the circuit :

10A

33.3A

3.33A

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Explanation

$(4) I = \frac{V}{\sqrt{R^{2} + ω^{2} L^{2}}} I = \frac{220}{\sqrt{400 + 2 π^{2} x 2500 x 4 x 10^{- 2}}} I = \frac{220}{\sqrt{400 + 16 π^{2} x 25}} = 3.33 A$

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A step-down transformer is connected to 2400 volts line and 80 amperes of current is found to flow in output load. The ratio of the turns in primary and secondary coil is 20 : 1. If transformer efficiency is 100%, then the current flowing in primary coil will be

1600 A

20 A

4 A

1.5 A

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Explanation

$\frac{N_{s}}{N_{p}} = \frac{V_{s}}{V_{p}} \Rightarrow \frac{1}{20} = \frac{V_{s}}{2400} \Rightarrow V_{s} = 120 V$

For 100% efficiency $V_{s} i_{s} = V_{p} i_{p}$

$\Rightarrow 120 \times 80 = 2400 i_{p} \Rightarrow i_{p} = 4 A$

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The potential difference V and the current i flowing through an instrument in an ac circuit of frequency f are given by $V = 5 \cos ω t$ volts and I = 2 sin ωt amperes (where ω = 2πf). The power dissipated in the instrument is

Zero

10 W

5 W

2.5 W

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Explanation

$V = 5 \cos ω t = 5 \sin (ω t + \frac{π}{2})$ and $i = 2 \sin ω t$

Power $= V_{r . m . s .} \times i_{r . m . s .} \times \cos ϕ$ = 0

(Since $ϕ = \frac{π}{2}$ , therefore $\cos ϕ = \cos \frac{π}{2} = 0$ )

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In an ac circuit, V and I are given by V = 100 sin (100 t) volts, $I = 100 \sin (100 t + \frac{π}{3}) m A$ . The power dissipated in circuit is

10⁴ watt

10 watt

2.5 watt

5 watt

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Explanation

$P = V_{r . m . s .} \times i_{r . m . s .} \times \cos ϕ$

$= \frac{100}{\sqrt{2}} \times \frac{100 \times 10^{- 3}}{\sqrt{2}} \times \cos \frac{π}{3}$

$= \frac{10^{4} \times 10^{- 3}}{2} \times \frac{1}{2} = \frac{10}{4} = 2 .5 w a t t$

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The resistance of a coil for dc is in ohms. In ac, the impedance:

Will remain the same

Will increase

Will decrease

Will be zero

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Explanation

The coil having inductance L beside the resistance R. Hence, for ac it’s effective resistance $\sqrt{R^{2} + X_{L}^{2}}$ will be larger than it’s resistance R for dc.

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A generator produces a voltage that is given by V = 240 sin 120 t, where t is in seconds. The frequency and r.m.s. voltage are

60 Hz and 240 V

19 Hz and 120 V

19 Hz and 170 V

754 Hz and 70 V

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Explanation

$ν = \frac{ω}{2 π} = \frac{120 \times 7}{2 \times 22} = 19 H z$

$V_{r . m . s .} = \frac{240}{\sqrt{2}} = 120 \sqrt{2} \approx 170 V$

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The peak value of an alternating e.m.f. E is given by $E = E_{0} \cos ω t$ is 10 volts and its frequency is 50 Hz. At time $t = \frac{1}{600} s e c$ , the instantaneous e.m.f. is

10 V

$5 \sqrt{3} V$

5 V

1 V

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Explanation

$E = E_{0} \cos ω t = E_{0} \cos \frac{2 π t}{T}$

$= 10 \cos \frac{2 π \times 50 \times 1}{600} = 10 \cos \frac{π}{6} = 5 \sqrt{3} v o l t .$

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If a current I given by $I_{0} \sin (ω t - \frac{π}{2})$ flows in an ac circuit across which an ac potential of $E = E_{0} \sin ω t$ has been applied, then the power consumption P in the circuit will be

$P = \frac{E_{0} I_{0}}{\sqrt{2}}$

$P = \sqrt{2} E_{0} I_{0}$

$P = \frac{E_{0} I_{0}}{2}$

P = 0

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Explanation

Phase angle $ϕ = 90^{o}$ , so power $P = V i \cos ϕ = 0$

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