two cells of emf 2 volt and 1 volt have internal resistances 1 ohm and 2 ohm respectively. their positive terminals are joined by a wire of resistance 10 ohm and negative terminals by a wire of resistance 4 ohm . another wire of resistance 10 ohm is connected between the middle points of these wires . find the current through the cells and the p.d. across the 10 ohm wire
Answers
Answer:
Explanation:
Given: Two cells of emf 1.5 V and 2.0 V and internal resistance 2 ohm and 1 ohm respectively have their negative terminals joined by a wire of 6 ohm and positive terminals by another wire of 4 ohm. A third wire of 8 ohm connects mid points of these two wires
To find the potential difference of the third wire
Solution:
This circuit diagram is as shown in figure. Let I
1
and I
2
be current given out by each cell.
Applying Kirchhoff's second law to closed mesh PQRS we have,
2I
1
+2I
1
+(I
1
+I
2
)8+3I
1
=1.5
or 15I
1
+8I
2
=1.5........(1)
Applying Kirchhoff's law to closed mesh RSNMR, we have
8(I
1
+I
2
)+3I
2
+1I
2
+2I
2
=2
or 8I
1
+14I
2
=2
or 4I
1
+7I
2
=1.................(2)
Multiplying (1) by 7 and (2) by 8, we have
105I
1
+56I
2
=10.5...........(3)
32I
1
+56I
2
=8...........(4)
Subtracting, (4) from (3) we get, 73I
1
=2.5
So, I
1
=
73
2.5
=
146
5
A
Putting this value of I
1
in (2),
4×
146
5
+7I
2
=1
or 7I
2
=1−
146
20
=
146
126
I
2
=
7
146
126
=
146
18
A
Thus, current through 8 ohms resistance
=I
1
+I
2
=
146
5
+
146
18
=
146
23
A
which is required value of current.
Also P.D across 8 ohms wire
= current × resistance
=
146
23
×8=
146
184
volts
So the required potential difference =1.26V