two cells of emf 3v and 4V and if internal resistance 1 ohm and 2 ohms respectively are connected in parallel so as to send current in the same direction through an external resistance of 5 ohms calculate the current through all the branches of the network
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0
Answer:
r
eff
E
eff
=
r
1
E
1
+
r
2
E
2
r
eff
=
r
1
+r
2
r
1
r
2
=
3+2
(3)(2)
=6/5Ω
6/5
E
eff
=
3
6
+
2
4
6
5
E
eff
=2+2=4
E
eff
=
5
24
I=
5
6
+8
5
24
=
5
46
5
24
=
46
24
I
8
=
46
24
×84
=
23
96
V
Explanation:
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