# Sinusoidal AC

## Trending Questions

**Q.**The equation of an alternating voltage is V =100 sin 100 π t volt. Its peak value and frequency are

- 50V, 100Hz
- 100 V, 50 Hz
- 100 V, 200 Hz
- 200 V, 100 Hz

**Q.**A resistance of 20 ohms is connected to a source of an alternating potential V=220 sin(100 πt). The time taken by the current to change from its peak value to r.m.s value is

- 0.25 sec
- 25×10−3sec
- 2.5×10−3sec
- 0.2 sec

**Q.**The peak value of an alternating e.m.f. E is given by E=E0cos ωt is 10 volts and its frequency is 50 Hz. At time t=1600sec, the instantaneous e.m.f. is

- 10 V
- 5√3V
- 5 V
- 1 V

**Q.**

The voltage of an ac source varies with time according to the equation V=100 sin 100 π t cos 100 π t where t is in seconds and V is in volts. Then

- The peak voltage of the source is 100 volts
- The peak voltage of the source is 50 volts
- The frequency of the source is 50 Hz
- The peak voltage of the source is 100√2 volts

**Q.**An ac voltage is represented by E=220√2 cos(50π)t. The current becomes zero

- 25
- 50
- 75
- 100

**Q.**The peak value of an alternating emf is given by E=E0cos(wt) is 10 volt and its frequency is 50 Hz. At a time t=(1600) second, the instantaneous value of the emf is:

- 1V
- 5V
- 10V
- 5√3V

**Q.**An ac generator produced an output voltage E = 170 sin 377 t volts, where t is in seconds. The frequency of ac voltage is

- 50 Hz
- 110 Hz
- 60 Hz
- 230 Hz

**Q.**Voltage and current in an ac circuit are given by V=5sin(100πt−π6) and I=4sin(100πt+π6)

- Voltage leads the current by 30∘
- Current leads the voltage by 30∘
- Current leads the voltage by 60∘
- Voltage leads the current by 60∘

**Q.**An electric lamp is connected to 220 V, 50 Hz supply. Then the peak value of voltage is

- 210 V
- 311 V
- 320 V
- 211 V

**Q.**Voltage of an AC source is given by V= 200sin(100π t).cos(100π t) where t is in seconds and V is in volts. Then the peak voltage is

volts.

- 50
- 100
- 200
- 400

**Q.**The voltage of an ac supply varies with time (t) as V = 120sin100π tcos100π t The maximum voltage and frequency respectively are

- 60 volts, 100 Hz
- 120 volts, 100 Hz
- volts , 100 Hz
- 60 volts, 200 Hz

**Q.**In an ac circuit, the current is given by i=4sin(100πt+30∘) ampere. The first time when the current becomes maximum (after t=0) at t is equal to -

**Q.**RMS current in a circuit of 50 Hz frequency is 5 A. Value of current 1300 seconds after its peak value is

- 5√32 A
- 2.5 A
- 5√2 A
- 5√2 A

**Q.**The voltage of an ac source varies with time according to the equation V = 200sin(100π t).cos(100π t) where t is in seconds and V is in volts. Then

- The peak voltage of the source is 100 volts
- The peak voltage of the source is 50 volts
- The frequency of the source is 50 Hz
- The peak voltage of the source is 100 / √2 volts

**Q.**

An ac voltage is represented by E=220√2 cos(50π)t How many times will the current becomes zero in 1 s?

50 times

100 times

30 times

25 times

**Q.**An ac source is rated at 220V, 50 Hz. The time taken for voltage to change from its peak value to zero is

- 5×10−3sec
- 0.02 sec
- 5 sec
- 50 sec

**Q.**What determines the frequency of a.c. produced in a generator?

- The number of rotations of the coil in one second.
- the speed of rotation of the coil
- both A & B
- None of the above

**Q.**The voltage of an ac supply varies with time (t) as V=120 sin 100 π t cos 100 π t. The maximum voltage and frequency respectively are

- 120 volts, 100 Hz
- 60 volts, 200 Hz
- 60 volts, 100 Hz
- 120√2 volts, 100 Hz

**Q.**If the phase difference between the emf and the current in an AC circuit is f then the RMS value of wattless current will be

- Irmscosf
- Irmssinf
- Irmstanf/2
- 0

**Q.**Match the following

Currentsr.m.s values(1) x0 sin ω t(i) x0(2) x0 sin ω t cos ω t(ii) x0√2(3) x0 sin ω t+x0 cos ω t(iii) x0(2√2)

- 1. (i), 2. (ii), 3. (iii)
- 1. (ii), 2. (iii), 3. (i)
- 1. (i), 2. (iii), 3. (ii)
- None of these

**Q.**

In the circuit given in the figure, switch S is closed at position 1 for long time. Find the total heat generated in resistor of resistance 2r0, when the switch S is open at position 1 and closed at position 2.

C0E204

C0E202

C0E20

C0E203

**Q.**An ac generator produced an output voltage E = 170 sin 377 t volts, where t is in seconds. The frequency of ac voltage is

- 50 Hz
- 110 Hz
- 60 Hz
- 230 Hz

**Q.**Four wires of identical length, diameters and of the same material are stretched on a sonometre wire. If the ratio of their tensions is 1 : 4 : 9 : 16, then the ratio of their fundamental frequencies are:

- 16 : 9 : 4 : 1
- 1 : 2 : 3 : 4
- 4 : 3 : 2 : 1
- 1 : 4 : 2 : 16

**Q.**A resistance of 20Ω is connected to a source of an alternating potential V=220sin(100 πt). The time taken by the current to change from the peak value to rms value, is

- 0.2 s
- 0.25 s
- 2.5×10−2 s
- 2.5×10−3 s

**Q.**Voltage and current in an ac circuit are given by V=5sin(100πt−π6) and I=4sin(100πt+π6)

- Voltage leads the current by 30∘
- Current leads the voltage by 30∘
- Current leads the voltage by 60∘
- Voltage leads the current by 60∘

**Q.**An electric lamp is connected to 220 V, 50 Hz supply. Then the peak value of voltage is

- 210 V
- 211 V
- 311 V
- 320 V

**Q.**The equation of an alternating current is I=50√2sin(400πt)A, then the frequency and the root mean square value of the current respectively are

- 200Hz, 50A
- 400Hz, 50√2A
- 200Hz, 50√2A
- 500Hz, 200A

**Q.**A resistance of 20 ohms is connected to a source of an alternating potential V=220 sin(100 πt). The time taken by the current to change from its peak value to r.m.s value is

- 0.2 sec
- 0.25 sec
- 25×10−3sec
- 2.5×10−3sec

**Q.**The instantaneous current from an a.c source is I=5 sin314 t. What is the (i) peak value (ii) r.m.s. value (iii) frequency of current?

**Q.**An alternating current is given by the equation i=i1cosωt+i2sinωt. The r.m.s. current is given by:

- 1√2(i1+i2)
- 1√2(i1+i2)2
- 1√2(i21+i22)1/2
- 12(i21+i22)1/2