1
Question
In an RLC series circuit shown in the figure, the readinf of voltmeter V1 and V2 are 100 V and 120 V, respectively. The source voltage is 130 V. For this situation, mark out the correct statement(s).
In an RLC series circuit shown in the figure, the readinf of voltmeter V1 and V2 are 100 V and 120 V, respectively. The source voltage is 130 V. For this situation, mark out the correct statement(s).
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Solution
The correct options are
A Voltage across resistor, inductor and capacitor are 50 V, 50√3 V and 120+50√3 V respectively
C Pwer factor of the circuit is 5/13
D The circuit is capacitive in nature
Given,
V1=100 V, ( it reads between inductor and resistor)
V2=120 V, (it reads between inductor and capacitor)
Source V=130 V
Let the potential difference across resistor, inductor and capacitor is VR, VL, VC respectively.
this voltage can be represented on the phasor diagram.
V2=V2R+(VL−VC)2(1)
Case 1: As given voltage across inductor and resistor is 100 V as we know the phase difference between them is π2. So,
V2R+V2L=1002(2)
Case 2: As given voltage across inductor and capacitor is 120 V as we know the phase difference between them is $π. So these can be directly subtracted,
|VL−VC|=120(3)
Note: this can be VC−VL thats why we are using mode for this case we assume VL>VC
From equation 1 and 3
1302=V2R−1202
V2R=2500
VR=50 V
Now, by using VR in equation 2
502+V2L=1002
VL=50√3 V
Substitute VL in equation 3
50√3−VC=120
VC=−120+50√3 V
So, VC is coming negative so it means our assumption is wrong so VC>VL
VC−VL=120
VC=(120+50√3) V
Now for power factor,
cosϕ=RZ=50130=513
But directly we also know that Z=130 V. So without this phasor diagram also we can directly calulate power factor.
As a potential difference across the capacitor is more than the potential difference across the inductor so it's a capacitive circuit. So option D is also correct
A Voltage across resistor, inductor and capacitor are 50 V, 50√3 V and 120+50√3 V respectively
C Pwer factor of the circuit is 5/13
D The circuit is capacitive in nature
Given,
V1=100 V, ( it reads between inductor and resistor)
V2=120 V, (it reads between inductor and capacitor)
Source V=130 V
Let the potential difference across resistor, inductor and capacitor is VR, VL, VC respectively.
this voltage can be represented on the phasor diagram.
V2=V2R+(VL−VC)2(1)
Case 1: As given voltage across inductor and resistor is 100 V as we know the phase difference between them is π2. So,
V2R+V2L=1002(2)
Case 2: As given voltage across inductor and capacitor is 120 V as we know the phase difference between them is $π. So these can be directly subtracted,
|VL−VC|=120(3)
Note: this can be VC−VL thats why we are using mode for this case we assume VL>VC
From equation 1 and 3
1302=V2R−1202
V2R=2500
VR=50 V
Now, by using VR in equation 2
502+V2L=1002
VL=50√3 V
Substitute VL in equation 3
50√3−VC=120
VC=−120+50√3 V
So, VC is coming negative so it means our assumption is wrong so VC>VL
VC−VL=120
VC=(120+50√3) V
Now for power factor,
cosϕ=RZ=50130=513
But directly we also know that Z=130 V. So without this phasor diagram also we can directly calulate power factor.
As a potential difference across the capacitor is more than the potential difference across the inductor so it's a capacitive circuit. So option D is also correct
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