Ⓒ A cell of emf 2 volt and internal reestance of 1/3-2 ts connected to two resistances of 10-2 and 20-2 joined in paralled, find current throgh each resistance.
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Answer:
Let R
1
and R
2
be the two resistance.
When the resistances are connected in series R
s
=R
1
+R
2
Current in the circuit is
I=
R+r
ϵ
i.e.,
5
2
=
R
s
+0
2
i.e., ⇒R
s
=R
1
+R
2
=5....(1)
When two resistances are connected in parallel
R
p
1
=
R
1
1
+
R
2
1
=
R
1
R
2
R
1
+R
2
=
R
1
R
2
5
(∵from(1))
R
p
=
5
R
1
R
2
Current in the circuit is
I=
R
p
+r
ϵ
3
5
=
(
5
R
1
R
2
)+0
2
=
R
1
R
2
10
R
1
R
2
=6....(2)
Using the relation (R
1
−R
2
)
2
=(R
1
+R
2
)
2
−4R
1
R
2
(R
1
−R
2
)
2
=(5)
2
−4(6)=25−24=1
∴R
1
−R
2
=1....(3)
Adding equations (1) and (3), we get
2R
1
=6 or R
1
=3Ω
R
2
=5−R
2
=5−3=2Ω.
Explanation:
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