Q: A series LCR circuit is connected to a 45 sin ( ω t ) Volt source. The resonant angular frequency of the circuit is 10 5 rad s − 1 and current amplitude at resonance is

Resonant angular frequency is given by, 1 L C = 10 5 1 50 × 10 - 3 C = 10 5 ⇒ C = 2 × 10 - 9 F Given: V = 45 sin ω t . Therefore, V 0 = 45 . Now, I 0 = V 0 R = 45 R . . . ii Inductive reactance, X L = ω L = 8 × 10 4 × 50 × 10 - 3 = 4000 Ω . and capacitive reactance, X C = 1 ω C = 1 8 &#21

Q: Consider an LC circuit, with inductance L = 0 . 1 H and capacitance C = 10 - 3 F , kept on a plane. The area of the circuit is 1 m 2 . It is placed in a constant magnetic field of

Emf induced in the circuit is E = d ϕ d t = d d t B 0 + β t A = β × A = 0 . 04 V So the circuit can be rearranged as Using Kirchhoff's law we can write E = L d i d t + q C L d i d t = E - q C For maximum current, d i d t = 0 , Or q = C E Using work energy theorem, q E = 1 2 L i 2 + q 2 2 C At i max , C E E = 1 2 L i m a x 2 + E 2 C 2 2 C Or, i max = C L E So, i max = 10 - 3 × 0 . 04 0 . 1 × 10 - 3 = 4 mA

Q: In a circuit, a metal filament lamp is connected in series with a capacitor of capacitance C F across a 200 ~V , 50 ~Hz supply. The power consumed by the lamp is 500 ~W while the voltage drop across i

Given, P = 500 W V R = 100 V f = 50 Hz V r m s = 200 V P = V R 2 R ⇒ R = V R 2 P ⇒ R = 100 2 500 ∴ R = 20 Ω Now current in the circuit, i = i r m s = voltage across resistor resistance i = 100 20 i = 5 A Phasor diagram of the given circuit for voltage, Now, V 2 = V C 2 + V R 2 ⇒ 200 2 = V C 2 + 100 2 ⇒ V C = 100 3 V Now, V C = 100 3 V = I X C ⇒ X C

Q: In a circuit, a metal filament lamp is connected in series with a capacitor of capacitance C F across a 200 ~V , 50 ~Hz supply. The power consumed by the lamp is 500 ~W while the voltage drop across i

Given, P = 500 W V R = 100 V f = 50 Hz V r m s = 200 V P = V R 2 R ⇒ R = V R 2 P ⇒ R = 100 2 500 ∴ R = 20 Ω Now current in the circuit, i = i r m s = voltage across resistor resistance i = 100 20 i = 5 A Phasor diagram of the given circuit for voltage, Now, V 2 = V C 2 + V R 2 ⇒ 200 2 = V C 2 + 100 2 ⇒ V C = 100 3 V Now, V C = 100 3 V = I X C ⇒ X C

Question 1

Consider a circuit consisting of a capacitor of capacitance C and a coil with N turns per unit length, cross sectional area S and length d, where d^2 S. There is another coil of length d/2, cross sect

Accepted Answer

Let the larger coil be coil 1 and the smaller coil be coil 2. The self-inductance of the larger coil is given by: L_1 = _0 n_1^2 A_1 l_1 = _0 N^2 S d = L The self-inductance of the smaller coil is: L_2 = _0 n_2^2 A_2 l_2 = _0 (2N)^2 ( S 2 ) ( d 2 ) = _0 (4N^2) ( Sd 4 ) = _0 N^2 S d = L The mutual inductance between the two coils is: M = _0 n_1 n_2 A_ common l_ common Since the smaller coil is completely inside the larger coil, the common area is S/2 and the common length is d/2. M = _0 (N) (2N)

Question 2

A circuit with an electrical load having impedance Z is connected with an AC source as shown in the diagram. The source voltage varies in time as V(t)=300 (400 t) V , where t is time in s. List-I show

Accepted Answer

For P i= V R =10 400 t (3) For Q aligned & X_L= L=400 100 10^ -3 =40 \ & Z=50 \ & . i= 300 50 (400 t-53^ ) [current will lag by ^ -1 X_L R ] (5) aligned For R aligned & X _ C = 10^6 400 50 =50 and X _ L =400 25 10^ -3 =10 \ & Z =50 \ & i = 300 50 (400 t +53^ ) [ Current will lead by ^ -1 X _ C - X _ L R ] (2) aligned For S i = 300 60 (400 t ) X _ L = X _ C Resonance (1)

Question 3

The circuit shown in the figure contains an inductor L, a capacitor C_0, a resistor R_0 and an ideal battery. The circuit also contains two keys K _1 and K _2. Initially, both the keys are open and th

Accepted Answer

(P) When K_1 is closed current in R_0 is I_1 At t =0; circuit will be aligned & I _1=0 \ & P (1) aligned (Q) After long time inductor behave as a wire so I _2 aligned & I _2= 20 5 =4 ~A \ & Q (3) aligned (R) When K _2 is closed and K _1 open aligned & _0= 1 LC \ & _0= 1 25 10^ -3 10 10^ -6 = 1 5 10^ -4 \ & _0=2 10^3 rad / s \ & _0=2 kilo-radian / s \ & R (2) aligned (S) Now K _2 is closed and K _1 open aligned & 1 2 LI _2^2= 1 2 CV _0^2 \ & 25 10^ -3 (4)^2=10 10^ -6 V _0^2 \ & ~V _0^2=25

Question 4

In a circuit shown in the figure, the capacitor C  is initially uncharged and the key K is open. In this condition, a current of 1 &#160; A flows through the 1 &#160; &#937; resistor. The key is

Accepted Answer

From second branch, we can see the potential drop across the branch is  5 + 1 &#215; 1 = 6 &#160; V . Now for the first branch, we can write 15 - I R = 6 &#160; &#160; &#160; . . . 1  and from third branch, we can write I 1 &#215; 3 = 6 &#8658; I 1 = 2 &#160; A Now from Kirchoff's junction rule, I = I 1 + 1 = 3 Now, from equation(1), we can write 15 - 3 R = 6 &#8658; R = 3 &#937; All three branches are in parallel, therefore we can write equivalent resistance as: 1 R eq = 1 3

Question 5

A series LCR circuit is connected to a 45 &#160; sin ( &#969; t ) Volt source. The resonant angular frequency of the circuit is 10 5 &#160; rad &#160; s &#8722; 1 and current amplitude at resonance is

Accepted Answer

Resonant angular frequency is given by,  1 L C = 10 5 1 50 &#215; 10 - 3 C = 10 5 &#8658; C = 2 &#215; 10 - 9 &#160; F Given:  V = 45 sin &#969; t . Therefore,  V 0 = 45 . Now,  I 0 = V 0 R = 45 R &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; . . . ii Inductive reactance,  X L = &#969; L = 8 &#215; 10 4 &#215; 50 &#215; 10 - 3 = 4000 &#160; &#937; . and capacitive reactance,  X C = 1 &#969; C = 1 8 &#21

Question 6

Consider an LC circuit, with inductance L = 0 . 1 &#160; H and capacitance C = 10 - 3 &#160; F , kept on a plane. The area of the circuit is 1 &#160; m 2 . It is placed in a constant magnetic field of

Accepted Answer

Emf induced in the circuit is E = d &#981; d t = d d t B 0 + &#946; t A = &#946; &#215; A = 0 . 04 &#160; V So the circuit can be rearranged as Using Kirchhoff's law we can write E = L d i d t + q C L d i d t = E - q C For maximum current,  d i d t = 0 ,  Or  q = C E Using work energy theorem, q E = 1 2 L i 2 + q 2 2 C At  i max ,  C E E = 1 2 L i m a x 2 + E 2 C 2 2 C Or,  i max = C L E So,  i max = 10 - 3 &#215; 0 . 04 0 . 1 &#215; 10 - 3 = 4 &#160; mA

Question 7

In a circuit, a metal filament lamp is connected in series with a capacitor of capacitance C F across a 200 ~V , 50 ~Hz supply. The power consumed by the lamp is 500 ~W while the voltage drop across i

Accepted Answer

Given,  P = 500 &#160; W V R = 100 &#160; V f = 50 &#160; Hz V r m s = 200 &#160; V P = V R 2 R &#8658; R = V R 2 P &#8658; R = 100 2 500 &#8756; &#160; &#160; R = 20 &#160; &#937; Now current in the circuit, i = i r m s = voltage &#160; across &#160; resistor resistance i = 100 20 i = 5 &#160; A Phasor diagram of the given circuit for voltage,   Now,  V 2 = V C 2 + V R 2 &#8658; 200 2 = V C 2 + 100 2 &#8658; V C = 100 3 &#160; V Now,  V C = 100 3 &#160; V = I X C &#8658; X C

Question 8

In a circuit, a metal filament lamp is connected in series with a capacitor of capacitance C F across a 200 ~V , 50 ~Hz supply. The power consumed by the lamp is 500 ~W while the voltage drop across i

Accepted Answer

Given,  P = 500 &#160; W V R = 100 &#160; V f = 50 &#160; Hz V r m s = 200 &#160; V P = V R 2 R &#8658; R = V R 2 P &#8658; R = 100 2 500 &#8756; &#160; &#160; R = 20 &#160; &#937; Now current in the circuit, i = i r m s = voltage &#160; across &#160; resistor resistance i = 100 20 i = 5 &#160; A Phasor diagram of the given circuit for voltage,   Now,  V 2 = V C 2 + V R 2 &#8658; 200 2 = V C 2 + 100 2 &#8658; V C = 100 3 &#160; V Now,  V C = 100 3 &#160; V = I X C &#8658; X C

Alternating Current — Previous Year Questions & Solutions