Any Elec engineers got an answer for this?
Does anyone know how to do this
A circuit having a resistance of 12 ohm, an indutance of 96mH and a capacitance of 52mF in series, is connected across a 220V, 50Hz supply. Calculate the following:
Total Impedance
current flowing through the circuit
the voltage across each component
the phase difference between the current and the supply voltage
The power factor of the circuit
Answer:
hi
i will answer ure question,,,,,,,,,
firstly ,
note what u are given in the question
R = 12 ohm
L = 96 * 10^-3
C = 52 * 10^-3
V = 220 V
w (frequency) = 50 Hz
==============================...
Total Impedence is denoted by Z
Z = √(XL - XC)^2 + R2
here,
XL is inductive reactance
XL = LW
XL = 96*10^-3*50
Xl = 4.8 ohms
similarly...
XC is capacitive reactance..
XC = 1/CW
XC = 1/52*10^-3*50
XC = 0.3846 ohms
Z = √(XL - XC)^2 + R^2
Z = √ (4.4153)^2 + 144
Z = 12.7863 (by calculator)
so the TOTAL IMPEDENCE IS Z = 12.7863 Ohms
==============================...
ANSWER-2)
Let I be the total current flowing thru the circuit....
V = I * Z
I = V / Z
I = 220/12.7863
I = 17.2059 ampere
so answer is I = 17.2059 ampere
==============================...
ANSWER - 3)
to find the voltage accross the resisitor
we use..
V= I * R
V= 17.2 * 12
V= 206.4 volts
to find the voltage accross a n inductor,,,,,,,,,
we use..
V = I * XL
V=17.2 * 4.8
V= 82.56 volts
to find the volatfe accross a capacitor...
we use...
V= I * XC
V = 17.2 * 0.3846
V = 6.61512 volts
==============================...
ANSWER -4 )
tan (theeta) = (XL - X C) /R
tan (theeta) = 4.4153/12
tan ( theeta) = 0.3679
theeta = tan-1 (0.3679)
this is the phase difference..k
==============================...
ANSWER-5)
Power factor = R / Z
phi is the power factor..
phi = 12/12.7863
phi = 0.9385
this is the power factor......
==============================...
Hope this is the best and most appropriate answer...
all the best....
FunQA.com will not be there to help in your job interview. Grab a book and a calculator.
Given: V=220, f=50hz, L=96millihenry, C=52millifarad
Find: Total Z=impedance
I= current
VL=voltage across inductance
VC=voltage across capacitor
VR=voltage across resistor
theta= phase angle
cos theta= power factor
Formulae:
XL=2pifL, XC=1/2pifC
w=2pif
Z=[(XL-XC)^2+R^2]^0.5
VL=I(XL), VC=I(XC), VR=IR , I=V/Z
Computation:
w=2pif=2(3.1416)50=314.16
XL=wL=314.16(96^-3)=30.159 ohms
XC=1/wC=1/314.16(52^-3)=0.0612 ohms
Z=(30.159-0.0612)^2+12^2]^0.5
=(906.01+144)^.5= 32.4 ohms, total impedance
I=total current=V/Z=220/32.4= 9.876 amps
VL=IXL= 9.876(30.159)= 297.86 volts across inductor
VC=IXC= 9.876(0.06)= 0.592 volts across capacitor
VR=IR= 9.876(12)= 118.512 volts across resistor
phase angle= arctan= (VL-VC)/VR
= (297.86-0.592)/118.512
=2.508
theta= tan^-1=2.508
phase angle=theta= 68.26 deg, voltage to current lag
power factor=cos theta=cos68.26 deg
=0.37 or 37%
hope this is clear enough, good luck
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A circuit having a resistance of 12 ohm, an indutance of 96mH and a capacitance of 52mF in series, is connected across a 220V, 50Hz supply. Calculate the following:
Total Impedance
current flowing through the circuit
the voltage across each component
the phase difference between the current and the supply voltage
The power factor of the circuit
Answer:
hi
i will answer ure question,,,,,,,,,
firstly ,
note what u are given in the question
R = 12 ohm
L = 96 * 10^-3
C = 52 * 10^-3
V = 220 V
w (frequency) = 50 Hz
==============================...
Total Impedence is denoted by Z
Z = √(XL - XC)^2 + R2
here,
XL is inductive reactance
XL = LW
XL = 96*10^-3*50
Xl = 4.8 ohms
similarly...
XC is capacitive reactance..
XC = 1/CW
XC = 1/52*10^-3*50
XC = 0.3846 ohms
Z = √(XL - XC)^2 + R^2
Z = √ (4.4153)^2 + 144
Z = 12.7863 (by calculator)
so the TOTAL IMPEDENCE IS Z = 12.7863 Ohms
==============================...
ANSWER-2)
Let I be the total current flowing thru the circuit....
V = I * Z
I = V / Z
I = 220/12.7863
I = 17.2059 ampere
so answer is I = 17.2059 ampere
==============================...
ANSWER - 3)
to find the voltage accross the resisitor
we use..
V= I * R
V= 17.2 * 12
V= 206.4 volts
to find the voltage accross a n inductor,,,,,,,,,
we use..
V = I * XL
V=17.2 * 4.8
V= 82.56 volts
to find the volatfe accross a capacitor...
we use...
V= I * XC
V = 17.2 * 0.3846
V = 6.61512 volts
==============================...
ANSWER -4 )
tan (theeta) = (XL - X C) /R
tan (theeta) = 4.4153/12
tan ( theeta) = 0.3679
theeta = tan-1 (0.3679)
this is the phase difference..k
==============================...
ANSWER-5)
Power factor = R / Z
phi is the power factor..
phi = 12/12.7863
phi = 0.9385
this is the power factor......
==============================...
Hope this is the best and most appropriate answer...
all the best....
FunQA.com will not be there to help in your job interview. Grab a book and a calculator.
Given: V=220, f=50hz, L=96millihenry, C=52millifarad
Find: Total Z=impedance
I= current
VL=voltage across inductance
VC=voltage across capacitor
VR=voltage across resistor
theta= phase angle
cos theta= power factor
Formulae:
XL=2pifL, XC=1/2pifC
w=2pif
Z=[(XL-XC)^2+R^2]^0.5
VL=I(XL), VC=I(XC), VR=IR , I=V/Z
Computation:
w=2pif=2(3.1416)50=314.16
XL=wL=314.16(96^-3)=30.159 ohms
XC=1/wC=1/314.16(52^-3)=0.0612 ohms
Z=(30.159-0.0612)^2+12^2]^0.5
=(906.01+144)^.5= 32.4 ohms, total impedance
I=total current=V/Z=220/32.4= 9.876 amps
VL=IXL= 9.876(30.159)= 297.86 volts across inductor
VC=IXC= 9.876(0.06)= 0.592 volts across capacitor
VR=IR= 9.876(12)= 118.512 volts across resistor
phase angle= arctan= (VL-VC)/VR
= (297.86-0.592)/118.512
=2.508
theta= tan^-1=2.508
phase angle=theta= 68.26 deg, voltage to current lag
power factor=cos theta=cos68.26 deg
=0.37 or 37%
hope this is clear enough, good luck
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