Alternating Current - Free MCQ Practice Test with solutions, EmSAT Achieve

For a positive half cycle, time period t = T/2

Then average e.m.f. during the positive half cycle of an A.C is
E_av (half cycle) = E/t
From equation 1 and 3

→ Equation for current and voltage for an A.C circuit can be written as:


Where, i and V are instantaneous values of current and voltage.

i₀ and V₀ are peak values of current and voltage.

ω = angular frequency in rad/sec

f = frequency in Hz

T = time period in sec.
Now, i_rms = i₀/√2
⇒ i₀ = i_rms × √2 = 5√2 A
∴ at t = 1/300 sec, and f = 50 Hz the current i will be,
⇒i = i₀ sin (2πf)t = (5√2) sin (2π×50) (1/300) 

So, the correct answer is option (2).

• DC voltmeter can not measure AC voltages.
• When used in AC circuits, DC Voltmeter gives zero reading because the average value of alternating voltage over a full cycle is zero.
• So, the reading of the voltmeter will be 0 volts, as a DC voltmeter cannot be used for measuring AC potential difference as it is current direction oriented.

• In a dc ammeter, a coil is free to rotate in the magnetic field of a fixed magnet.
• If an alternating current is passed through such a coil, the torque will reverse its direction each time the current changes direction, and the average value of the torque will be zero.
• Hence, the average current value over a complete cycle is zero and the alternating current cannot be measured by the DC ammeter.

• The electric current flows in two ways: Alternating current and Direct Current.
  • Direct current flows only in one direction.

• Alternating current: The electric current whose direction changes periodically is called electric current.

• Alternating current reverses its direction periodically.
• It also changes its magnitude periodically because of the induced electromagnetic force.
• For Alternating current both magnitude and direction change. The frequency of the alternating current in the Indian power supply is 50 Hz. The time period is 1/50 = 20 msec.

From the above explanation, we can see that for a pure capacitance circuit if AC voltage of v = v_m sinωt then the driven current in the capacitor will be 

I = i_m sin(ωt + ϕ) 

Now as mentioned for pure inductor phase angle will be lagging by 90° or π/2.

i.e., ϕ = π/2  ⇒ i = i_m sin (ωt + π/2)

Given -  I = 2cos(ωt+θ)

• The value of I will be maximum when cos(ωt+θ) is maximum,
• The maximum value of cos(ωt+θ) = 1

Therefore,
⇒ I_o = 2A
So,

Hence, option 1 is correct.

From the above explanation, we can see that power decapitated by any circuit can be expressed as 

Whereas for A.C current the above expression can be expressed as 

Hence option 2 is correct among all

• The peak value of the alternating current is the maximum value attained by the alternating current.
• The RMS value of alternating current is given as,
  
• The mean value of alternating current is zero. So the mean value is minimum. Hence, option 3 is correct.

• The DC ammeter measures the mean value of current and the mean value of the AC current is zero, so we cannot measure the AC current by the DC ammeter.
• Hence, option 2 is correct.