Sinusoidal AC
Trending Questions
Q.
An alternating voltage is represented as . The average value of voltage over one cycle will be
Q. The voltage of domestic ac is 220 volt. What does this represent
- Mean voltage
- Peak voltage
- Root mean voltage
- Root mean square voltage
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 supply varies with time (t) as V = 120sin100π tcos100π t The maximum voltage and frequency respectively are
- 120 volts, 100 Hz
- 120√2 volts , 100 Hz
- 60 volts, 200 Hz
- 60 volts, 100 Hz
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)
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. The voltage of domestic ac is 220 volt. What does this represent
- Mean voltage
- Peak voltage
- Root mean voltage
- Root mean square voltage
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. 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. An alternating voltage is represented as E=20 sin 300 t. The average value of voltage over one cycle will be
- Zero
- 10volt
- 20√2volt
- 20√2volt
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. 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.
volts.
- 50
- 100
- 200
- 400
Q. An ac voltage is represented by E=220√2 cos(50π)t. The current becomes zero times in 1 sec.
- 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
- 5√3V
- 10V
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 peak voltage of the source is 100√2 volts
- The frequency of the source is 50 Hz
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. 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.
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