Q 1 :    

An alternating current at any instant is given by i=[6+56sin(100πt+π3)]A. The rms value of the current is _______ A    [2024]



(8)     If i=a+b sin(ωt+θ)

         irms=a2+b22

        irms=36+562

        =36+28=64=8A

 



Q 2 :    

An alternating voltage V(t)=220 sin 100πt volt is applied to a purely resistive load of 50Ω. The time taken for the current to rise from half of the peak value to the peak value is                [2024]

  • 5 ms

     

  • 3.3 ms

     

  • 7.2 ms

     

  • 2.2 ms

     

(2)

V(t)=220sin(100πt)

I=VsR=22050sin(100πt)

I=225sin(100πt)

cosθ=I02I0=12θ=60°=π3

Using phasor, t=π/3100π

t=1300 sec=103 msec

t=3.3 msec



Q 3 :    

An alternating current is given by

I=IA sin ωt+IB cos ωt

The r.m.s. current will be          [2025]

  • IA2+IB2

     

  • IA2+IB22

     

  • IA2+IB22

     

  • |IA+IB|2

     

(3)

I=IA sin ωt+IB cos ωt

So, IRMS=IA2+IB22



Q 4 :    

An electric bulb rated as 100 W – 220 V is connected to an ac source of rms voltage 220 V. The peak value of current through the bulb is:          [2025]

  • 0.64 A

     

  • 0.45 A

     

  • 2.2 A

     

  • 0.32 A

     

(1)

RB=V2P=220×220100=484Ω

Irms=VrmsR=220484=1022=511A

I0=5211=0.64 A



Q 5 :    

An alternating current is represented by the equation, i=1002 sin (100 πt) ampere. The RMS value of current and the frequency of the given alternating current are          [2025]

  • 1002 A, 100 Hz

     

  • 1002 A, 100 Hz

     

  • 100 A, 50 Hz

     

  • 502 A, 50 Hz

     

(3)

i=1002 sin (100 πt)

irms=10022=100 A

f=ω2π=100π2π=50 Hz



Q 6 :    

An ac current is represented as

i=52+10 cos (650 πt+π6) Amp

The r.m.s. value of the current is          [2025]

  • 20 Amp

     

  • 100 Amp

     

  • 10 Amp

     

  • 52 Amp

     

(3)

i=52+10 cos (650 πt+π6)

i2=50+100 cos2 (650 πt+π6)+(2)(52)(10) cos (650πt+π6)

<i2>=50+1002+0

<i2>=100

<i>=10 Amp