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JEE Main 27-Aug-2021 Shift 2 Solved Paper
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© examsnet.com
Question : 27
Total: 90
An AC circuit has an inductor and a resistor of resistance R in series, such that
X
L
=
3
R
. Now, a capacitor is added in series such that
X
C
=
2
R
. The ratio of new power factor with the old power factor of the circuit is
√
5
:
x
The valueof x is.
[27 Aug 2021 Shift 2]
Your Answer:
Validate
Solution:
Given, resistance of resistor
=
R
Inductance,
X
L
=
3
R
Capacitance
X
C
=
2
R
As we know that,
Power factor,
cos
φ
=
R
Z
where,
Z
is impedance,
Z
=
√
R
2
+
(
X
L
−
X
C
)
2
For case I, an inductance connected in series with resistance
cos
φ
1
=
R
Z
1
=
R
√
R
2
+
(
X
L
−
0
)
2
=
R
√
R
2
+
(
3
R
)
2
=
R
R
√
10
=
1
√
10
⋅
⋅
⋅
⋅
⋅
⋅
⋅
(
i
)
For case II, A capacitor is also connected in series with resistance
∴
cos
φ
2
=
R
Z
2
⇒
cos
φ
2
=
R
√
R
2
+
(
3
R
−
2
R
)
2
=
R
√
R
2
+
R
2
=
R
R
√
2
=
1
√
2
.
.
.
(ii)
Now, ratio of Eqs. (ii) and (i), we get
cos
φ
2
cos
φ
1
=
1
√
2
1
√
10
=
√
10
2
=
√
5
:
1
∴
x
=
1
© examsnet.com
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