two cells of e.m.f. 1.5V and 2v and internal resistance 2ohm and 1ohm respectively have their negative terminals joint by a wire of 6ohm and positive terminals by a wire of 4ohm resistance.a third resistance wire of 8ohm connects middle point of these wires.using kirchhoff 's law.find the potential difference at the end of this third wire?

Asked by nishu | 3rd Jul, 2013, 03:02: PM

Expert Answer:

This circuit  diagram is as shown in figure. Let I1 and I2 be current given out by each cell.
Applying Kirchoff's second law to closed mesh PQRS we have,
2I1 + 2I1+ (I1 + I2)8 + 3I1 = 1.5
or 15 I1 + 8 I2 = 1.5 ........(1)
Applying kirchhoff's law to closed mesh RSNMR, we have
8(I1 + I2) + 3I2 + 1I2 + 2I2 = 2
or 8I1 + 14I2 = 2
or 4I1 + 7I2 = 1.................(2)
Multiplying (1)  by 7 and (2) by 8, we have
105I1 + 56I2 = 10.5 ...........(3)
32I1 + 56I2 = 8 ...........(4)
Subtracting, (4) from (3) we get, 73I1 = 2.5
So, I1  = 2.5/73 = 5/146 amp.
Putting this value of I1 in (2),
4 x 5/146 + 7I2 = 1
or 7I2 = 1 - 20/146 = 126/146
I2 = (126/146)/7 = 18/146 amp.
Thus, current through 8 ohms resistance
= I1 + I2 = 5/146 + 18/146
= 23/146 amp.
which is required value of current.
Also P.D across 8 ohms wire
= current x resistance
= (23/146) x 8 = 184/146 volts
= 1.26 volts

Answered by  | 5th Jul, 2013, 12:58: PM

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