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AP EAMCET Engineering 2018 Apr 22 Shift 2 Paper
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© examsnet.com
Question : 114
Total: 160
An inductor and a resistor are connected in series to an ac source. The current in the circuit is 500 mA if the applied ac voltage is
8
√
2
V at a frequency of
175
Ï€
Hz and the current in the circuit is 400 mA if the same ac voltage at a frequency of
225
Ï€
Hz is applied. The values of the inductance and resistance are respectively
60 mH, 71 Ω
√
60
mH, 71 Ω
√
60
mH,
√
71
Ω
60 mH,
√
71
Ω
Validate
Solution:
Given info:
I
=
500
×
10
−
3
A
ω
1
=
175
Ï€
×
2
Ï€
‌
rad
∕
s
=
350
‌
rad
∕
s
V
1
=
8
√
2
Write the expression of current for an L – R circuit.
I
=
V
Z
=
V
√
R
2
+
L
2
ω
2
R
2
+
L
2
ω
2
=
(
V
I
)
2
R
2
+
L
2
(
350
)
2
=
(
8
√
2
500
×
10
−
3
)
2
R
2
+
L
2
(
350
)
2
=
512
...... (1)
And,
R
2
+
L
2
(
350
)
2
=
800
...... (1)
By solving Equation (1) and (2) we get,
R =
√
71
Ω and L= 60 mH,
© examsnet.com
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