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Alternating Current
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Section:
Physics
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© examsnet.com
Question : 1 of 13
Marks:
+1
,
-0
A circuit with an electrical load having impedance
Z
Z
Z
is connected with an AC source as shown in the diagram. The source voltage varies in time as
V
(
t
)
=
300
sin
(
400
t
)
V
V(t)=300 \sin(400 t)\,\text{V}
V
(
t
)
=
300
sin
(
400
t
)
V
, where
t
t
t
is time in s. List-I shows various options for the load. The possible currents
i
(
t
)
i(t)
i
(
t
)
in the circuit as a function of time are given in List-II.
Choose the option that describes the correct match between the entries in List-I to those in List-II.
List-I
List-II
(P)
(1)
(Q)
(2)
(R)
(3)
(S)
(4)
(5)
[JEE Adv 2025 P1]
P
⟶
3
,
Q
⟶
5
,
R
⟶
2
,
S
⟶
1
P \longrightarrow 3,\; Q \longrightarrow 5,\; R \longrightarrow 2,\; S \longrightarrow 1
P
⟶
3
,
Q
⟶
5
,
R
⟶
2
,
S
⟶
1
P
⟶
1
,
Q
⟶
5
,
R
⟶
2
,
S
⟶
3
P \longrightarrow 1,\; Q \longrightarrow 5,\; R \longrightarrow 2,\; S \longrightarrow 3
P
⟶
1
,
Q
⟶
5
,
R
⟶
2
,
S
⟶
3
P
⟶
3
,
Q
⟶
4
,
R
⟶
2
,
S
⟶
1
P \longrightarrow 3,\; Q \longrightarrow 4,\; R \longrightarrow 2,\; S \longrightarrow 1
P
⟶
3
,
Q
⟶
4
,
R
⟶
2
,
S
⟶
1
P
⟶
1
,
Q
⟶
4
,
R
⟶
2
,
S
⟶
5
P \longrightarrow 1,\; Q \longrightarrow 4,\; R \longrightarrow 2,\; S \longrightarrow 5
P
⟶
1
,
Q
⟶
4
,
R
⟶
2
,
S
⟶
5
Validate
Solution:
V
(
t
)
=
300
sin
(
400
t
)
V(t)=300 \sin(400 t)
V
(
t
)
=
300
sin
(
400
t
)
(P)
30S
\;\;\text{ 30S }\;
30S
⇒
i
0
=
300
30
=
10
A
\;\Rightarrow \;\; i_0=\frac{300}{30}=10\,\text{A}
⇒
i
0
=
30
300
=
10
A
and
ϕ
=
0
\phi=0
ϕ
=
0
⇒
(
P
)
⟶
(
3
)
\Rightarrow \;\; (P) \longrightarrow (3)
⇒
(
P
)
⟶
(
3
)
(Q)
⇒
Z
=
R
2
+
(
ω
L
)
2
\;\Rightarrow \;\; Z=\sqrt{R^2+(\omega L)^2}
⇒
Z
=
R
2
+
(
ω
L
)
2
=
(
30
)
2
+
(
400
×
100
×
1
0
−
3
)
2
=
50
Ω
\;\; =\sqrt{(30)^2+(400 \times 100 \times 10^{-3})^2}=50\,\Omega
=
(
30
)
2
+
(
400
×
100
×
1
0
−
3
)
2
=
50
Ω
∴
i
0
=
300
50
=
6
A
\; \therefore \;\; i_0=\frac{300}{50}=6\,\text{A}
∴
i
0
=
50
300
=
6
A
φ
=
tan
−
1
(
ω
L
R
)
=
5
3
∘
(
lag
)
\; \;\; \varphi=\tan^{-1}\left(\frac{\omega L}{R}\right)=53^{\circ}\,(\text{lag})
φ
=
tan
−
1
(
R
ω
L
)
=
5
3
∘
(
lag
)
(
Q
)
⟶
(
5
)
\; \;\; (Q) \longrightarrow (5)
(
Q
)
⟶
(
5
)
(R)
Z
=
(
30
)
2
+
(
1
400
×
50
×
1
0
−
6
−
400
×
25
×
1
0
−
3
)
2
Z=\sqrt{(30)^2+\left(\frac{1}{400 \times 50 \times 10^{-6}}-400 \times 25 \times 10^{-3}\right)^2}
Z
=
(
30
)
2
+
(
400
×
50
×
1
0
−
6
1
−
400
×
25
×
1
0
−
3
)
2
=
(
30
)
2
+
(
40
)
2
=
50
Ω
=\sqrt{(30)^2+(40)^2}=50\,\Omega
=
(
30
)
2
+
(
40
)
2
=
50
Ω
∴
R
⟶
(
2
)
\therefore \;\; R \longrightarrow (2)
∴
R
⟶
(
2
)
(S)
Z
=
(
60
)
2
+
(
1
50
×
1
0
−
6
×
400
−
125
×
1
0
−
3
×
400
)
Z=\sqrt{(60)^2+\left(\frac{1}{50 \times 10^{-6} \times 400}-125 \times 10^{-3} \times 400\right)}
Z
=
(
60
)
2
+
(
50
×
1
0
−
6
×
400
1
−
125
×
1
0
−
3
×
400
)
=
(
60
)
2
+
(
50
−
50
)
2
=\sqrt{(60)^2+(50-50)^2}
=
(
60
)
2
+
(
50
−
50
)
2
=
60
Ω
=60\,\Omega
=
60
Ω
∴
i
0
=
300
60
=
5
A
\therefore \;\; i_0=\frac{300}{60}=5\,\text{A}
∴
i
0
=
60
300
=
5
A
(
S
)
⟶
1
\;(S) \longrightarrow 1
(
S
)
⟶
1
© examsnet.com
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