In an AC circuit with resistance R = 1 Ω and oscillating potential v = vm sinωt, where vm = 1 V and ω = 2π rad/s, the value of average current in time t ∈ [0 s, 0.5 s] is,

  

CONCEPT:

• The oscillating potential of an AC circuit is given by

⇒ v = vm sinωt

where 'vm' is the amplitude of the oscillating potential difference and 'ω' is the angular frequency

• The oscillating current in an AC circuit is given by

\(⇒ i = \frac{v_m}{R} sin ω t\)

where 'i' is the amplitude of the oscillating current

since 'vm' and 'R' are constants, 

⇒ i = im sinωt

where \( i_m = \frac{v_m}{R} \)

​

EXPLANATION:

Given: resistance R = 1 Ω, oscillating potential v = vm sinωt, where vm = 1 V and ω = 2π rad/s

• The oscillating potential of an AC circuit is given by

⇒ v = vm sinωt

where 'vm' is the amplitude of the oscillating potential difference and 'ω' is the angular frequency

• The oscillating current in an AC circuit is given by

\(⇒ i = \frac{v_m}{R} sin ω t\)

where 'i' is the amplitude of the oscillating current

• The average current over a half sinusoidal cycle is

\(⇒ i_{avg} = \frac {1}{t}\int_0^{t} \frac{v_m}{R} sin ω t dt\)

\(⇒ i_{avg} = \frac {1}{0.5}\int_0^{0.5} \frac{1}{1} sin (2\pi t) dt = -\frac {1}{0.5\times 2\pi} (cos (2\pi \times 0.5) -cos 0)=\frac {2}{\pi}A\)

• ​Therefore option 1 is correct.