Test: Alternating Current - 2

Effective voltage is also known as root mean square value of voltage of alternating current in an AC circuit.
we know, E_rms=E_peak/ √2 
where E_rms  denotes the rms voltage and E_peak peak voltage in an ac circuit.
here, peak voltage , E_peak =423 volts
so, effective voltage , E_rms  =423/√2  ≈ 300 volts
hence, effective voltage is 300 volts
The correct answer is option C.

i=4sin(100πt+ π/6)
For I to be maximum
di/dt=0
therefore,4x100π(cos(100πt+π/6)=0
therefore, ωs(100πt+ π/6)=0
for the first time cosθ=0 at π/2
⇒ 100πt+ π/6= π/2
⇒ 100πt= π/3
⇒ πt= π/300
T=1/300

Instantaneous current i=2sin(100πt+π/3)A
At (t=0),  i(0)=2sin(π/3)=2×√3/2​​=√3​A
The current at the beginning will be √3​ A
 

I=10sin(314t)
dt      [ω=314, T=2π/​ω=2π/​314, T=π/​157]

I_avg=(2/T)x10 [-cos314/314]^T/2
=-20/(Tx314)[cos{314x(π/2x157)}-cos0]
       =20/3.14
       =6.37A

Given,
Instantaneous current, i=3+6sinωt
i_rms​=√[(1/T​)0T(3+6sinωt)2dt​]
i_rms​=√(1/T) 0T​(32+62sin2ωt+36sinωt) dt​
i_rms​=√(1/T​) [0T32dt+∫0T​62{(1/2) ​– (cos2ωt/2​) dt+∫0T​(36sinωt)dt]​
i_rms​=√9+(36/2) ​+0​=3√3​A 
Hence, RMS value of current is 3√3​A

l=500sin(100πt)
⟹ω=100π
ω=2πf
⟹2πf=100π
f=50Hz
 

We know: Vrms​=​Vo​​/√2
 
Substituting values (Here RMS Voltage is 220 V)
We get: Maximum V (or amplitude) =311V
 

For alternating current average value is taken for half cycles only,
Let, I=I0sinωt where ω=2π/T

=[(I₀/ω) cosωt]_o^T/2 x 2/T
I_avg=2I_o/ π

Given peak potential is 200v
so, amplitude is 200
and frequency is 50Hz
so angular frequency is 2.πx50 = 314/sec
so the value is
E = 200sin(314t)
Hence option B is the correct answer.

I=(I1)cosωt+(I₂)sinωt
(I²)_mean=I₁²cos²ωt+I₂²sin²ωt+I₁I₂cosωtsinωt
I₁²×(1/2)×i₂²×(1/2)+2_I1I₂×0
I_rms=√(i²)_mean=√(I₁²+I₂²/2)=(1/√2)( I₁²+I₂²)^1/2

E= ε_ocos(ωt+π/3) –1
I=I₀sin(ωt)=I₀cos{(π/2)-cot)
 =I0cos(ωt+(- π/2)) —2
By,1 and 2,
Phase difference between E and I,
=ϕ_E-V_I
=(ωt+π/3)- (ωt+(- π/2))
= π/3+ π/2
=5π/6

We know that,
X_L= ωL
Or, X_L =2πfL   [ ω can be written as, 2πf]
Initially,
X_L1=2πf₁L₁=1000Ω
For the second case,
F  and L both are doubled, so,
X_L2 =2π2f₁2L₁ =4x2πf₁L₁
                        =4000 Ω

R=1W
X_L=1000W
X_C=1000W
Z= Z= √ (R²+(X_L-X_C)²)
  =√(1²+(1000-1000)²)
Z=1W
    I=V/Z=100/1
I=1000A
 

L =0.14
E=100sin(100t)
⇒X_L​=w_L​=100×0.1
=10Ω
⇒i_o​=E_o/XL​=100/10​=10A
⇒i=i_o​sin(100t− π/2​)
⇒i=−i_o​sin[(π/2)​−100]
So we get
⇒i=−10cos(100t)A
therefore, option D is correct.

Given,
Reactance of a coil at 50 Hz, X_L=100ohm
If frequency becomes 150 Hz, let the reactance be X_L’
As, X_L= ωL
          =2πfL
X_L=2π(50)L……(1)
              X_L’=2π(150)L…….(2)
Dividing 2 by 1
X_L’/X_L=2π(150)L/2π(50)L
=>X_L’/100=150/50
=>X_L’=150x100/50
         =300ohm
   X_L’=300ohm
Hence reactance of the coil when frequency at 150Hz is 300ohm, option (B).
 

Given, R=50W, L=(20/π)H and C=(5/π)mF
Voltage supply, V=230V.
v=50Hz
Since, X_L= ω=2πvL=2πx50x(20/π)=X_L=2000 Ω
And, X_C= 1/ω_C=1/2πvL=1/[2πx50x(5/π)x10^-6]=(1000/500) x10³
So, impedance, Z=√[(X_L-X_C)²+R²]
Z=√[(2000-2000)²+50²]
Z=50Ω
Hence, Option B is correct.

V=ZI ->Impedance
I=V/Z
V_R=IR=VR/Z=>R=(V_R/V)Z
V_L=IX_L=VX_L/Z=>X_L=(V_L/V)Z
Similarly, X_L=(V_C/V)Z
Impedance, Z²=R²+(X_L-X_C)²
=>Z²=(V_R²/V²)Z² +[{(V_L/V)Z}-{(V_C/V)Z}]²
V=√[V_R²+(V_L-V_C)² ]
On substituting, V_L =3V_R   , V_C=2V_R
=>V=√2 V_R

Power factor (cosϕ)=R/Z₁
For first case, power factor=0.866
Or, R/ Z₁=0.866
Or,Z₁=1.154R
For second case, power factor=>0.5
Cosϕ₂=>R/Z₂=>0.5, Z₂=2R.
% increase=>[(Z₂-Z₁)/Z₁] x100
So, [{(2R)-(1.154R)}/(1.154R)]x100=73.2%
So the correct answer is option A.

v₀=1/2π√(LC)
⇒In this question,
L’=L+25% of L
=L+(L/4)=5L/4
And C’=C-20% of C
=C-(C/5)=4C/5
Hence, v₀=1/2π√(L’C’)
=1/2π√[(5L/4) x (4C/5)]
=1/2π√(LC)=v₀
The answer is option C, remains unchanged.
 

X_net(reactance)=XL-XC=0, =>X_L=X_C
ω=1/√(L_C)
Z=√(R² + (X_net)²)

In an LCR ac circuit, 
Impedance, Z=R+j(XL​−XC​) . . . . . . ..  .(1)
At the resonance, inductive reactance is equal to the capacitive reactance. 
XL​=XC​
From equation (1),
Z=R
The impedance is independent of the frequency term.
On increasing the frequency, the impedance will remain unchanged.
The correct option is B.
 

The voltage can be written as,
V=5(sinωt+(π/2​))
The angle between the voltage and the current is 90O.
The power dissipated in the circuit is given as,
P=V_rms​I_rms​cosϕ
=V_rms​I_rms​cos90O
=0
Thus, the power dissipated in the circuit is 0.
 

Choke Coil=inductor
Φ=90o
P=VIcos Φ
 =VIcos90o
 =0

First bulb= v=220v
P=500w
P₁= (v²/R₁)
R₁= V²/P₁= 220x200/500 ohms
For second bulb:
v=220 v
P=300W
R2= v2/p=220x200/300
By taking ratios:
R₁:R₂= 220x200x 300: 500 x220x200 = 3:5

Resistance= (voltage)²/Power
Resistance of 40 W bulb= (200)²/40=1000 ohm
Resistance of 60 W bulb=(200)2/60=666.67 ohm
Resistance of 100 w bulb=(200)2/100=400 ohm
Total resistance= 1000+666.67+400=2066.67 ohm (addition because series circuit)
Now, we calculate current flowing in the bulbs (as we know this is series circuit so same current will flow in the circuit)
Current (I)= Voltage/total resistance=200/2066.67=0.0967 A
Now we find the power consumed by each bulb,
power consumed by 40 W bulb=I2R=(0.0967)2x1000=9.35 W
Power consumed by 60 W bulb=I2xR=(0.0967)2x666.67=6.23 W
Power consumed by 100 W bulb=I2R=(0.0967)2x400=3.74 W
From the above results, we can say that power consumed by a 40 W bulb is more as compared to other bulbs so 40 W glows brighter than other bulbs.
 

The alternating current changes its direction and magnitude, it becomes zero twice in a complete cycle. In one half cycle current flows in one direction and changes its direction for other half cycle. So, it is not possible by direct current meters to read such a high fluctuation in a second due to inertia and hence, no deflection accounted. Hence, alternating current can not be measured by direct current meters, because the average value of current for the complete cycle is zero.