Q: In the circuit shown in the figure, the AC source gives a voltage V=20 (2000 t) neglecting source resistance, the voltmeter and ammeter reading will be

Given, R_1=8 aligned R_2 & =2 \\ L & =5 mH \\ C & =50 F aligned and V_0=20 (2000 t) Impedance, Z= R^2+ (X_L-X_C )^2 As, here, X_L= L=2000 5 10^ -3 =10 Similarly, X_L= 1 C = 1 2000 50 10^ -6 =10 X_L=X_C, Hence, i_ = V_0 Z = 20 (8+2) =2 ~A Hence, i_ rms = i_ 2 = 2 2 =1.41 ~A and aligned V & =R_2 i_ rms =1.41 2 \\ & =282 ~V aligned

Q: For an RLC circuit driven with voltage of amplitude V_ m and frequency _ 0 = 1 LC the current exhibits resonance the quality factor, Q is given by

Total angular momentum of the system initially $ L _ i = I _ 1 w + I _ 2 (0)= I _ 1 w $ Total angular momentum of the system finally $ L _ f = ( I _ 1 + I _ 2 ) w _ 2 $ According to conservation of angular momentum i.e. $ L _ i = L _ f $ $ aligned & I _ 1 = ( I _ 1 + I _ 2 ) _ 2 \\ & _ 2 = I _ 1 I _ 1 + I _ 2 aligned $

Question 1

Find the average value of current shown graphically from t=0 to t=2 ~s .

Accepted Answer

From i-t graph, area from t=0 to t=2 ~s , we get aligned & = 1 2 1 10+ 1 2 (2-1) 10 \ & =5+5=10 ~A \ Area current, i_ av & = Area (i-t) graph time interval = 10 2 =5 ~A aligned

Question 2

In an ac circuit, an inductor, a capacitor and a resistor are connected in series with X_L=R= X _ C . Impedance of this circuit is :

Accepted Answer

Impedance in LCR circuit Z= (X_L-X_C )^2+R^2 X_L=X_C=R Z=R

Question 3

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

Accepted Answer

Rising half to peak t=T / 6 t = 2 6 = 3 = 300 = 1 300 =3.3 ~ms Here =100 rad / s

Question 4

In the given circuit, rms value of current (Irms) through the resistor R is :

Accepted Answer

(a) Given, Capacitive reactance, X _ C =100 Inductive reactance, X _ L =200 Resistance, R=100 Impedance, Z = R ^2+ ( X _ L - X _ C )^2 Z= 100^2+(200-100)^2 =100 2 RMS value of current, i_ rms = V_ rms Z = 200 2 100 2 =2 ~A

Question 5

In the circuit shown in the figure, the AC source gives a voltage V=20 (2000 t) neglecting source resistance, the voltmeter and ammeter reading will be

Accepted Answer

Given, R_1=8 aligned R_2 & =2 \ L & =5 mH \ C & =50 F aligned and V_0=20 (2000 t) Impedance, Z= R^2+ (X_L-X_C )^2 As, here, X_L= L=2000 5 10^ -3 =10 Similarly, X_L= 1 C = 1 2000 50 10^ -6 =10 X_L=X_C, Hence, i_ = V_0 Z = 20 (8+2) =2 ~A Hence, i_ rms = i_ 2 = 2 2 =1.41 ~A and aligned V & =R_2 i_ rms =1.41 2 \ & =282 ~V aligned

Question 6

An alternating voltage V = V _ 0 t is applied across a circuit. As a result, a current I = I _ 0 ( t - 2 ) flows in it. The power consumed per cycle is

Accepted Answer

The phase angle between voltage V and current I is 2 . Therefore, power factor = ( 2 )=0. Hence the power consumed is zero.

Question 7

For an RLC circuit driven with voltage of amplitude V_ m and frequency _ 0 = 1 LC the current exhibits resonance the quality factor, Q is given by

Accepted Answer

Total angular momentum of the system initially $ L _ i = I _ 1 w + I _ 2 (0)= I _ 1 w $ Total angular momentum of the system finally $ L _ f = ( I _ 1 + I _ 2 ) w _ 2 $ According to conservation of angular momentum i.e. $ L _ i = L _ f $ $ aligned & I _ 1 = ( I _ 1 + I _ 2 ) _ 2 \ & _ 2 = I _ 1 I _ 1 + I _ 2 aligned $

Alternating Current — Previous Year Questions & Solutions