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

Given:
𝑅
=
3.3
 
k
Ω
=
3300
 
Ω
R=3.3kΩ=3300Ω

𝐿
=
120
 
mH
=
0.12
 
H
L=120mH=0.12H

𝑓
=
2
 
kHz
=
2000
 
Hz
f=2kHz=2000Hz

𝑉
source
=
12
 
V (rms)
V
source
​
=12V (rms)

Step 1: Calculate the current through the resistor,
𝐼
𝑅
I
R
​
:
The current through a resistor is given by Ohm's Law:

𝐼
𝑅
=
𝑉
𝑅
I
R
​
=
R
V
​

Substitute the known values:

𝐼
𝑅
=
12
3300
=
0.003636
 
A
=
3.636
 
mA
I
R
​
=
3300
12
​
=0.003636A=3.636mA
Step 2: Calculate the inductive reactance,
𝑋
𝐿
X
L
​
:
The inductive reactance
𝑋
𝐿
X
L
​
is given by:

𝑋
𝐿
=
2
𝜋
𝑓
𝐿
X
L
​
=2πfL
Substitute the known values:

𝑋
𝐿
=
2
𝜋
×
2000
×
0.12
=
2
𝜋
×
240
=
1507.96
 
Ω
X
L
​
=2π×2000×0.12=2π×240=1507.96Ω
Step 3: Calculate the current through the inductor,
𝐼
𝐿
I
L
​
:
The current through the inductor is given by:

𝐼
𝐿
=
𝑉
𝑋
𝐿
I
L
​
=
X
L
​

V
​

Substitute the known values:

𝐼
𝐿
=
12
1507.96
=
0.00796
 
A
=
7.96
 
mA
I
L
​
=
1507.96
12
​
=0.00796A=7.96mA
Step 4: Find the total current using Kirchhoff’s Current Law (KCL):
Since the resistor and the inductor are in parallel, the total current
𝐼
total
I
total
​
is the vector sum of the currents through the resistor and inductor:

𝐼
total
=
𝐼
𝑅
2
+
𝐼
𝐿
2
I
total
​
=
I
R
2
​
+I
L
2
​

​

Substitute the values for
𝐼
𝑅
I
R
​
and
𝐼
𝐿
I
L
​
:

𝐼
total
=
(
3.636
)
2
+
(
7.96
)
2
=
13.21
+
63.36
=
76.57
=
8.74
 
mA
I
total
​
=
(3.636)
2
+(7.96)
2

​
=
13.21+63.36
​
=
76.57
​
=8.74mA

Admin answer is wrong plz correct it.
Correct answer is 36