Test: Alternating Current - 1 - JEE MCQ

Explanation:The standard voltage and frequency of alternating current supply in India is set to 220 V and 50Hz respectively by the government of India becausewhen power has to be transmitted from a power plant, the biggest challenge is to cut the transmission losses. For this purpose, the current value should be small and potential difference(Voltage) should be more. Also losses are minimal at 50 Hz/60 Hz frequency.

Resonance occurs when X_L = X_C and the imaginary part of the transfer function is zero. At resonance the impedance of the circuit is equal to the resistance value as Z = R.

I₀=250μA, v₀=3.6v .  v=1.6x10⁶ Hz
Here,
(Reactance of inductance) X_L=ωL
X_L=2πv X L
v₀/I₀=2πv x L
3.6/2.5x10^-4=2πx1.6x10^-6 x L
0.14x10^4-6=L
L=0.14x10^-2H
Now for v=16.0x10⁶Hz
XL=2πv X L
=2πx16x10⁶x14x10^-4
X_L=1407x10²Ω
Now,
v0=I0 x XL
3.6/1407x10²=I₀   [∵v0=kept constant.]
I₀=0.00256x10^-2
I₀=25.6μA

Explanation:By using the phenomenon of mutual induction, transformers allow us to easily change voltage of AC. This is necessary to cut down poer losses while supplying electricity to our homes

Since the current flowing through a series resonance circuit is the product of voltage divided by impedance, at resonance the impedance, Z is at its minimum value, ( =R ). Therefore, the circuit current at this frequency will be at its maximum value of V/R

Angular frequency of the ac signal w=2πν
∴ w=2π(50)=100π
Capacitive reactance X_c​=1/wC​
∴ X_c​=1/(100π×786×10⁻⁶)​=4Ω
Inductive reactance X_L​=w_L
∴ X_L​=100π×(25.48×10−3)=8Ω
Impedance of the circuit Z=√[R²+(X_L​−X_c​)²​]
∴ Z=√[3²+(8−4)²​]=5Ω
 Phase difference ϕ=tan⁻¹[(XL​−Xc​​)R]
Or ϕ=tan⁻¹((8−4​)/3)=tan⁻¹(4/3​)
⟹ ϕ=53.13o
Power factor cosϕ=cos53.13o=0.6
Power dissipated in the circuit
P=I_v²R
Now, Iv=I₀/√2=E₀/√2Z=283/(1.414×5)=40A
∴P=I_v²R=(40)²×3=4800 watt

The voltage in India is 220 volts, alternating at 50 cycles (Hertz) per second. This is the same as, or similar to, most countries in the world including Australia, Europe and the UK. However, it's different to the 110-120 volt electricity with 60 cycles per second that's used in the United States for small appliances.

Explanation:Alternating Current is the current in which the polarity of source continuously changes on a fixed frequency. So the positive and the negative terminals ‘alternate’

The Voltage Transformer can be thought of as an electrical component rather than an electronic component. A transformer basically is very simple static (or stationary) electro-magnetic passive electrical device that works on the principle of Faraday’s law of induction by converting electrical energy from one value to another.
 
The transformer does this by linking together two or more electrical circuits using a common oscillating magnetic circuit which is produced by the transformer itself. A transformer operates on the principles of “electromagnetic induction”, in the form of  Mutual Induction.
Mutual induction is the process by which a coil of wire magnetically induces a voltage into another coil located in close proximity to it. Then we can say that transformers work in the “magnetic domain”, and transformers get their name from the fact that they “transform” one voltage or current level into another.

At any time, t
energy stored in capacitor, EC=q²/2C
Total energy stored in C and L at any time t
E=E_C+E_L
E=(q²/2C)+(LI²/2)
If q=q₀sinωt
I=dq/dt=q₀cosωtω=ωq0cosωt
Put in (i), E=[(q₀sinωt)²/2C] + L(ωq₀cosωt)²/2
E=(q₀²/2C) sin²ωt+(q₀²/2)cos2ωt.ω²L
But ω²=12/LC
∴E=(q₀²/2C) sin²ωt+(q₀²/2) (L/LC) cos²ωt
=q₀²/2C(sin²ωt+cos²ωt)
=q₀²/2C= constant in time.
 

Changes(if required): The answer provided in the forum is wrong, The correct answer would be option B.
Solution:
Given,
V_R-rms =30.0V
V_C-rms=90.0V
V_L-rms=50.0V
To find the root mean square voltage across the source, we will use the formula,
V_rms=√[V_r²+(V_L-V_C)²]
=√[(30.0V)² +(50.0V-90.0V)²]
=50.0 V

Explanation:By convention phasor diagrams are made to show the rotation of phasor in counterclockwise direction with constant speed

 So the range is 88 - 198 pF

Explanation:In the first half of the cycle, power is supplied to the inductive circuit. In the next half the inductor dissipates the energy supplied by circuit.Thus overall no power is supplied to the circuit

Given    P=50W, V_rms​=220V,
by     P=V_rms^2​​/R
or      R=V_rms²/P​​= 220²/50=968Ω
The peak voltage of the source is,
E_rms=E₀/√2
=>E₀=Erms x √2=220√2=311.13V
Now, 
I_rms​=V_rms​​/R
=220/968​
=0.227A
Hence option B is correct.

Given,
P_avg=1200W
V_rms=120 Volts
A,
The resistance is,
R=V_rms²/P_avg
So,R=120²/1500=9.6ohm
B,
The rms current is,
I_rms=V_rms/R
So, I_rms=120/9.6
=>I_rms=12.5A
C,
The maximum instantaneous power is,
P_m=2P_avg
So,P_m=2(1500)
=>P_m=3000W.
Hence option D is the correct answer.
 

The resistance is R=200ohm the inductor is L=0.4 H, the capacitance is C=5  µF and the amplitude voltage is V=CV
The frequency depends on the inductance and the capacitance and it is given by,
f₀=1/ (2π√(LC))  
So, plug the values of L and C into equation 1 to get f₀
f₀=1/ (2π√(LC))=2/(2π√[0.4H)(5x10^-6)]=113Hz
for the current, we use ohm’s law to get the current,
I=V/R
Now, plug the values for V and R to get I
I=V/R=3V/200Ω=0.015A=15mA
 

Natural frequency of the circuit, f = 1/2π√(LC)
here, L = 20mH = 20 × 10^-3 H, C = 50μF = 50 × 10^-6 C
so, f = 1/2π√(20 × 10^-3 × 50 × 10^-6)
= 1/2π√(1000 × 10^-9)
= 1/2π√(10^-6)
= 1000/2π = 500/π Hz ≈ 159 Hz

Here V_p=220V V_s=110V
As we know the relation between V and n,
As,
V_e/V_s=n_p/ns ->220/110
Np/ns=2/1=2:1
Therefore, no. of turns in primary is greater than no. of turns in secondary,
Hence, it is a step-down transformer.

Given,
The resistance is R=200ohm the inductor is L=0.4 H, the capacitance is C=5  µF and the amplitude voltage is V=CV
The frequency depends on the inductance and the capacitance and it is given by,
f₀=1/ (2π√(LC))  (1)
So, plug the values of L and C into equation 1 to get f₀
f₀=1/ (2π√(LC))=2/(2π√[0.4H)(5x10-6)]=113Hz
 
The inductance reactance of the coil and could be calculated by,
X_L=ωL
Plug the values for L and ω to get XL
X_L= ωL=2π(113Hz)(0.4)=160ohm
 To get the capacitive reactance we use the formula,
X_C=1/ωC
Now, we plus the values for ω and C to get XC
X_C=1/ ωC=1/2π(113Hz)(5x10^-6)=500ohm
We use the inductor, the resistor and the capacitor, so the impedance of the circuit is given by the equation,
Z=√(R²+(X_L-XC)² )  (2)
Now, we use the values for XR,XL and R into equation (2) to get Z
Z= √ (R2+(XL-XC)2 )
   =√ ((200Ω)2+(160Ω-500Ω)2 )
   =394.5Ω
The impedance represents the total resistance in the circuit, so we can use the value in Ohm’s law to get the current amplitude I in the circuit,
I=V/Z
So, plus the values of V and Z to get I,
I=V/Z=3V/394.5Ω=0.0076A=7.61mA

In an inductor, current lags behind the input voltage by a phase difference of π/2.
Current and voltage are in the same phase in the resistor whereas current leads the voltage by π/2 in a capacitor.
So, the circuit must contain an inductor only.

When frequency decreases, capacitive reactance XC​=1/2πνC​ increases and hence impedance in the circuit Z=√(R2+XC2​)​ increases and so current I=V/Z​ decreases. As a result, the brightness of the bulb is reduced.
 

Given,
Capacitance, C=100μF=100x10^-6F
10^-4F
Resistance, R=40Ω
Rms voltage, E_v=110 volt
Peak voltage, E₀=√2 .E_v=√2 x 110V
Frequency of Ac supply, v=60Hz.,
ω=2πv=120π rad/s
peak current, I₀=?
a.   In RC,as
Z=√(R²+X_C²)= √[R²+(1/ ω²C²)]
Therefore,
I₀=E₀/√[R²+(1/ ω²C²)]
  =√2x100/√[1600+(1/120πx10^-4)²]
I₀=3.23amp
b.   In RC circuit voltage lags behind the current by phase angle Φ, where Φ is given by,
tanΦ=(1/ω)/R
=1/ωCR
=1/(120πx10-4x40)
=0.6628
⇒ Φ=tan^-1(0.6628) =33.5o
=(33.5π/180)rad
Time lag= Φ/ω
=33.5π/180x120π
=1.55x10^-3 sec.
=1.55ms
 

Given,
We are given the current I=1.5Am where the wire will break.
We are asked to determine the root mean square of the current Irms. The maximum current here represents the current that just after it the wire will break. The maximum value of the current is the amplitude of the current wave and it should be larger than the root mean square of the current. Using equation, we can get the I_rms in the form,
I_rms = I_max/√2
The term, 1/√2, times any factor represents the root mean square of this factor, Now, plug the value for Imax into equation 1 and get I_rms
I_rms=I_max/√2
     =1.5A/√2
     =1.06A.

Explanation:We know that in capacitor Current leads Voltage by 90degree. Over one complete cycle, in first quarter cycle Capacitor charges and next quarter cycle it's discharge. This will continue in next negative half cycle. So the NET POWER ABSORB IS ZERO.