In a series RL circuit, 12 V rms is measured across the resistor and 14 V rms is measured across the inductor. The peak value of the source voltage is
A. 18.4 V
B. 26.0 V
C. 2 V
D. 20 V
Answer: Option B
Solution (By Examveda Team)
$$\eqalign{ & {{\text{E}}_{{\text{rms}}}} = \sqrt {{\text{V}}_{\text{L}}^2} + {\text{V}}_{\text{R}}^2 \cr & \Rightarrow {{\text{E}}_{{\text{rms}}}} = \sqrt {{{12}^2}} + {14^2} = \sqrt {340{\text{V}}} \cr & {{\text{E}}_{{\text{max}}}} = {{\text{E}}_{{\text{rms}}}}\sqrt 2 \cr & {{\text{E}}_{{\text{max}}}} = \sqrt {680{\text{V}}} \cr & \Rightarrow {{\text{E}}_{{\text{max}}}} = 26.0\,{\text{V}} \cr} $$This Question Belongs to Electrical Engineering >> RL Circuits
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Comments (2)
Related Questions on RL Circuits
If the frequency is halved and the resistance is doubled, the impedance of a series RL circuit
A. Doubles
B. Halves
C. Remains constant
D. Cannot be determined without values
When the frequency is decreased, the impedance of a parallel RL circuit
A. Increases
B. Decreases
C. Remains constant
D. Is not a factor
Given:
π
=
3.3
β
k
Ξ©
=
3300
β
Ξ©
R=3.3kΞ©=3300Ξ©
πΏ
=
120
β
mH
=
0.12
β
H
L=120mH=0.12H
π
=
2
β
kHz
=
2000
β
Hz
f=2kHz=2000Hz
π
source
=
12
β
VΒ (rms)
V
source
β
=12VΒ (rms)
Step 1: Calculate the current through the resistor,
πΌ
π
I
R
β
:
The current through a resistor is given by Ohm's Law:
πΌ
π
=
π
π
I
R
β
=
R
V
β
Substitute the known values:
πΌ
π
=
12
3300
=
0.003636
β
A
=
3.636
β
mA
I
R
β
=
3300
12
β
=0.003636A=3.636mA
Step 2: Calculate the inductive reactance,
π
πΏ
X
L
β
:
The inductive reactance
π
πΏ
X
L
β
is given by:
π
πΏ
=
2
π
π
πΏ
X
L
β
=2ΟfL
Substitute the known values:
π
πΏ
=
2
π
Γ
2000
Γ
0.12
=
2
π
Γ
240
=
1507.96
β
Ξ©
X
L
β
=2ΟΓ2000Γ0.12=2ΟΓ240=1507.96Ξ©
Step 3: Calculate the current through the inductor,
πΌ
πΏ
I
L
β
:
The current through the inductor is given by:
πΌ
πΏ
=
π
π
πΏ
I
L
β
=
X
L
β
V
β
Substitute the known values:
πΌ
πΏ
=
12
1507.96
=
0.00796
β
A
=
7.96
β
mA
I
L
β
=
1507.96
12
β
=0.00796A=7.96mA
Step 4: Find the total current using Kirchhoffβs Current Law (KCL):
Since the resistor and the inductor are in parallel, the total current
πΌ
total
I
total
β
is the vector sum of the currents through the resistor and inductor:
πΌ
total
=
πΌ
π
2
+
πΌ
πΏ
2
I
total
β
=
I
R
2
β
+I
L
2
β
β
Substitute the values for
πΌ
π
I
R
β
and
πΌ
πΏ
I
L
β
:
πΌ
total
=
(
3.636
)
2
+
(
7.96
)
2
=
13.21
+
63.36
=
76.57
=
8.74
β
mA
I
total
β
=
(3.636)
2
+(7.96)
2
β
=
13.21+63.36
β
=
76.57
β
=8.74mA
Admin answer is wrong plz correct it.
Correct answer is 36