**Linked comprehensive Type**
A circuit consists of a capacitor of X

_{c}=30 ohm , a non-inductive resistor of 44 ohm and a coil of inductive resistance 90 ohm and resistance 36 ohm in series is connected to 200 V ,60 hz AC circuit

**Question 1** Find the Impedance (Z) of the circuit

a) 10 ohm

b) 100 ohm

c) 30 ohm

d) 40 ohm

**Question 2** Find the current in the circuit?

a) 1 A

b) 3 A

c) 2 A

d) 5 A

**Question 3** Find the Impedance of the coil

a) 97 ohm

b) 50 ohm

c) 90 ohm

d) None of these

**Question 4** Match the following

__Column A__
A) V

_{Coil}
B) V

_{Res}
C) V

_{Cap}
D) V

_{Coil} + V

_{Res} + V

_{Cap}
__Column B__
P) 88 V

Q) 194 V

R) 200 V

S) 60 V

T) No appropriate match

**Question 6** Match the following

__Column A__
A) Power dissipated in coil

B) Power dissipated in Resistor

C) Power loss of the circuit

__Column B__
P) 144 W

Q) 320 W

R) 176 W

S) No appropriate match

Solutions for Question 1-6
Given

X_{c}=30 ohm

R_{1}=44 ohm

X_{L}=90 ohm

R_{2}=36 ohm

Impedance (Z) of the circuit is given by

$Z=\sqrt {(R_1+ R_2)^2 + (X_L-X_C)^2}$

$Z=\sqrt {(44+ 36)^2 + (90-30)^2} =100 \Omega $

So Answer for (1) is b

Now current in the circuit is given by

$I=\frac {V}{Z}=200/100=2 A$

So Answer for (2) is c

Impedance of the coil us given by

$Z_c=\sqrt {R_2^2 + X_L^2}$

=97 ohm

So Answer for (3) is a

Now

V_{C}=IX_{C}=2*30=60 V

V_{R}=IR=2*44=88V

V_{L}=IZ_{C}=2*97=194 V

V_{Coil} + V_{Res} + V_{Cap}= 60 +88+194=342 V

So

A-> Q

B-> P

C-> S

D -> T

Now

Power dissipated is given by

P=I^{2}R

Power dissipated in Coil

P=(2)^{2}*36= 144 W

Power dissipated in Resistor

P= =(2)^{2}*44=176 W

Total power dissipated =144+176=320 W

**Question 7**:

An Alternating voltage (in volts) varies with time (in sec) as

$V=100 \sqrt {2} sin (100 \pi t)$

Match with column with above context in mind

__Column A__
A) Peak value of Voltage

B)Rms value of the voltage

C) Frequency of the voltage

D)Mean square value of the voltage

__Column B__
P) 100

Q) $10 \sqrt 2$

R) 50

S) 10

^{4}
Solution
An Alternating voltage varies with time as

$V=V_0 sin \omega t$

Given

$V=100 \sqrt {2} sin (100 \pi t)$

Comparing the two equations ,we have

$V_0=100 \sqrt{2} $ Volts

$\omega =100 \pi $ i.e $2 \ pi \nu =100 \pi $

Or frequency=50 Hz

$V_rms= \frac {V_0}{\sqrt 2} =100 Volt$

$V_ms = \ frac {V_0^2}{2} = 10^4 $

**Question 8**: An L-R circuit contains an inductor of inductance 10 Henry and resistance of 2 ohm. It is connected to 10 Volt battery

a) How long will it take for the current to reach the half the maximum value

b) Find the time constant

c) Find out the value of dI/dt at t =0

Solution
The Growth of current in an LR circuit is given by

**Question 9**:

An LCR circuit in series has following values

R=11 Ohm

X

_{C}=120 Ohm

X

_{L}= 120 ohm

The circuit is connected with 110 V ,60 Hz power source

__Match the column__
__Column I__
A) V

_{L}
B) V

_{C}
C) V

_{R}
D) $\sqrt { V_R^2 +(V_L -V_C)^2$

__Column II__
P) 1200V

Q) 110

R) 1100V

T) Data not sufficient

Solution
I=E/Z

Where

$Z=\sqrt {R^2 + (X_L - X_C}^2}$

Substituting all the values

Z=11 ohm

So current I =110/11=10 A

Now

V_{R}=IR=10*11=110 V

V_{L}=IX_{L}=10*120=1200 V

V_{C}=IX_{C}=10*120=1200 V

$\sqrt { V_R^2 + (V_L - V_C)^2 }=110 V$

**Linked comprehensive Type**
An circuit consists of a series combination of a 50mH inductor and a 20μF capacitor. The circuit is connected to AC supply of 220V and 50 Hz .

The circuit has the value of R=0

**Question 10** Which of the following is correct for the circuit?

a) I

_{0}=2.17 A ,I

_{rms}= 1.53 A

b) I

_{0}=5.1 A ,I

_{rms}= 3.6 A

c) V

_{0}=311 V ,V

_{rms}=200 V

\) None of these

**Question 11** which of the following is false

a) Voltage drop across inductor is 23.1 V

b) Voltage drop across capacitor is 243 V

c) Voltage drop across inductor is 243 V

d) Voltage drop across capacitor is 23.1 V

**Question 12** which of the following is true

a) Average power transferred to inductor is zero

b) Average power transferred to capacitor is zero

c) Average power absorbed by the circuit is zero

d) None of the above

Solutions for Question 10-12
The following quantities are given in the question

L=50mH=50X10^{-3} H

C=20μF=20X10^{-6}F

V_{rms}=220V

f=50hz or ω=2πf=100π rad/sec

Now

V 0 = 2 V rms = 311 V

The peak current is given by the below equation in LC circuit

I 0 = V 0 Z

Where

Z = 1 ωL − 1 ωC

Substituting the values from above

I_{0}=-2.17 A

So magnitude is

I_{0}=2.17 A

I_{rms}=I_{0}/ 2 =1.53 A

Voltage drop across inductor is given by

V_{L}=I_{rms}ωL=23.1 V

Voltage drop across capacitor is given by

V_{C}= I rms ωC =243 V

Since in inductor and capacitor, voltage and current are at right angle, no power will be transferred to them

Hence total power absorbed is also zero

Class 12 Maths
Class 12 Physics