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 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
- 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.
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)
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