Linked comprehensive Type
A circuit consists of a capacitor of Xc=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) VCoil
(B) VRes
(C) VCap
(D) VCoil + VRes + VCap 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
Xc=30 ohm
R1=44 ohm
XL=90 ohm
R2=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
VC=IXC=2*30=60 V
VR=IR=2*44=88V
VL=IZC=2*97=194 V
VCoil + VRes + VCap= 60 +88+194=342 V
So
A-> Q
B-> P
C-> S
D -> T
Now
Power dissipated is given by
P=I2R
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) 104 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
XC=120 Ohm
XL= 120 ohm
The circuit is connected with 110 V ,60 Hz power source Match the column Column I
(A) VL
(B) VC
(C) VR
(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
VR=IR=10*11=110 V
VL=IXL=10*120=1200 V
VC=IXC=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) I0=2.17 A ,Irms= 1.53 A
(b) I0=5.1 A ,Irms= 3.6 A
(c) V0=311 V ,Vrms=200 V
(d) 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=20X10-6F
Vrms=220V
f=50hz or ω=2πf=100π rad/sec
Now
$V_0 = \sqrt {V_rms} = 311$ V
The peak current is given by the below equation in LC circuit
$I_0 = \frac {V_0}{Z}$
Where
$Z = \omega L - \frac { 1}{ \omega C}$
Substituting the values from above
I0=-2.17 A
So magnitude is
I0=2.17 A
$I_{rms}=\frac {I_0}{ \sqrt 2} =1.53$ A
Voltage drop across inductor is given by
VL=IrmsωL=23.1 V
Voltage drop across capacitor is given by
$V_C= \frac {I_{rms}} {\omega C} =243$ 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
Linked comprehensive Type
The circuit is shown below.
Currently the switch is in position A and steady current is flowing to the circuit. Capacitor is uncharged. At t=0, switch is instanously moved to position B. Question 13 Find the current in the LC circuit for t≥ 0
Here $\omega = \frac {1} {\sqrt LC}$ Solution
Question 14 Find the charge q on the lower capacitor plate for t≥ 0 Solution
Question 15 Which of the following is true of the circuit?
(a) The electric oscillations of time period $2\pi \sqrt{LC}$ are produced in the LC circuit
(b) The total energy remains conserved and is equal to $\frac {1}{2}L \frac {V_0^2}{R}$
(c) The maximum current in the circuit is V0/R
(d) None of these Solution
$T=2\pi \omega =2 \pi \sqrt {LC}$
Total energy=Initial energy in the inductor
$E=\frac {1}{2} LI^2 =\frac {1}{2} L \frac {V_0^2}{R}$
Question 16
An LR circuit is connect to Voltage source V0. Which of the following is true at time equal to time constant
(a) The rate at which energy is being stored in Inductor is given by $\frac {.63 V_0^2}{R}$
(b) Power loss in the resistor is given by $\frac {.40 V_0^2}{R}$
(c) The rate of change of current is $\frac {.37 V_0^2}{R}$
(d) Current is 63% of the maximum value Solution
Question 17
A circuit is shown below in Figure
It is given
L=R2C
Which of the following is true
(a) The total current will instantly rise to its final value V0/R when switch is closed
(b) The current will zero in LR branch and have value V0/R in the LC circuit path at t=0
(c) The current will rise with time in the LR circuit and decrease with time in LC circuit
(d) Total current in the circuit will always be V0/R for t>0 Solution
Let IL denote the current in the RL branch
And Ic denote the current in RC Branch
Then Total current through battery
I=IL + IC
Now since the battery is across two branches in parallel the current through the RL branch is unaffected by the presence of the RC branch ,
Now it is given
L=R2C
Or
$\frac {L}{R}=RC$
So
$I=\frac {V_0}{R}$
So I is independent of time and equal to V0/R for t ≥0
Also (1) and (2)
The current will zero in LR branch and have value V0/R in the LC circuit path at t=0
The current will rise with time in the LR circuit and decrease with time in LC circuit
Linked comprehensive Type
A sinusoidal voltage of frequency f=60Hz and peak value 150V is applied to a series LR circuit
Where L=40mH and R=20ohm Question 18 Which one of the following is correct Question 19 At t=T/6, which one of the following is true
(a) The instantaneous current is 2.344 A
(b) The instantaneous voltage across inductor is 83.29 V
(c) The instantaneous voltage across resistor is 46.88 V
(d) The total voltage applied is 130 V Question 20 Which one of the following is true about the circuit?
(a) Average power into the circuit is 360 W
(b) Irms is 4.24 A
(c) Vrms is 106.07 V
(d) None of these Solutions for Questions 6-8
(a) Now from the formula’s
$T=\frac {1} {f}=\frac {1}{60} sec$
$\omega =2\pi f= 377rad/sec $
$X_L=\omega L=15.08 \Omega $
$Z=\sqrt {R^2 + X_L^2} =25.05 \Omega $
$\phi = tan ^-1 \frac {X_L){R}=37^0 $
Now max current
$I=\frac {V}{Z}=150/25.05=6A$
Max voltage across Resistor is
VR=IR=120V
Max Voltage across Inductor is
VL=IXL =90.5V
(b) At t=T/6
We have
$\omega t=\frac {\pi }{3}=60^0$
Instantaneous current is given by
$i=I sin (\omega t - \phi) =6 sin 23=2.344 A$
Instantaneous voltage across Inductor
$v=V_L cos (\omega t - \phi)=90.5 cos 23=83.29 V$
Instantaneous voltage across Resistor
vR = iR==46.88
Total voltage =vL + vR =130 V
(c)
From the formula’s
$I_rms= \frac {I}{\sqrt 2} =\frac {6}{\sqrt 2} =4.24 A$
$V_rms=\frac {V}{\sqrt 2} = 106.07 V$
Average power= (Irms )2 R=360 W
Question 21:
A coil of inductance .50H and resistance 100 ohm is connected to 240 V,50Hz AC supply
Which of the following is true
a) Irms in the coil is 1.28 A
b) The time lag between voltage maximum and current maximum is 3.2X10-3 s
c) Maximum Voltage is 339.36 V
d) None of these Solution
Given
Vrms=240 V ,L=.50H and R=100ohm
Now
Now for time lag,
Voltage of an ac source varies as
$V=V_0 cos \omega t$
And similarly current varies as
$I=I_0 cos (\omega t - \phi )$
Where
$tan \phi =\frac {\omega L}{R}$
Now for t=0,Voltage is maximum
I will be maximum when ωt-Φ=0 i.e when t=Φ/ω
So that time lag between voltage maximum and current maximum is
$\Delta t=\frac{ \phi }{ \omega }$
Now let us calculate Φ
$tan \phi =\frac {\omega L}{R} =\frac {2 \pi (50)(.5)}{100}=1.57$
Or
Φ=57.50=57.5Xπ/180 rad
So time lag
$\Delta t=\frac{ \phi }{ \omega } =3.2 x 10^-3 s$
Question 21 Consider a series RC circuit for which R = 2.0 × 106, C = 6.0 µF and E = 20 V. Find
(i) The time constant of the circuit,
(ii) The maximum charge on the capacitor after a switch in the circuit is closed, and
(iii) The current in the circuit at the instant just after the switch in the circuit is closed
Link Type comphrehension
A coil of resistance 15 Ohm and inductance .6 H is connected to a steady 120V power source.
The current starts from zero until it reaches it final state. $\frac{dI}{dt}$ is the rate of increase of current until it reaches steady state.
Question 22
Which equation will be correct until current reaches steady state
(a) $\frac{dI}{dt}=\frac{2-75I}{3}$
(b) $\frac{dI}{dt}=\frac{600-75I}{6}$
(c) $\frac{dI}{dt}=\frac{600-5I}{3}$
(d) $\frac{dI}{dt}=\frac{600-75I}{3}$ Solution
The effect driving voltage in the circuit will be the power minus the induced EMF in the inductor
So, Potential difference equation of the circuit will be
$120-.6\frac{dI}{dt}=15I$
Or
$\frac{dI}{dt}=\frac{600-75I}{3} $ ---(1)
Question 23 Match the Following Column A
(a) $\left(\frac{dI}{dt}\right)_{t=0}$
(b) $\left(\frac{dI}{dt}\right)$ when the current reaches 80% of the maximum values
(c) $\left(\frac{dI}{dt}\right) $ when the current reach the maximum value
(d) The current I when $\left(\frac{dI}{dt}\right)$ is 150 A/s
At t=0,I is essentially zero
So from equation (1)
$\frac{dI}{dt}=\frac{600-75I}{3}$
$\left(\frac{dI}{dt}\right)_{t=0}=\frac{600}{3}=200 $ A/s
Now current reaches it maximum value when current finally stop changing and induced emf becomes zero i.e when \frac{dI}{dt}=0
So maximum current
$0=\frac{600-75I}{3}$
Or
I=8 A
So 80% of current will be =.8*8=.64 A
So from equation (1)
$\frac{dI}{dt}=\frac{600-75I}{3}$
$\frac{dI}{dt}=\frac{600-75*.64}{3}=40$A/s
Now when $\frac{dI}{dt}=150$ A/s
So from equation (1)
