A 6-volt battery of negligible internal resistance is connected across a uniform wire AB of length 100cm. The positive terminal of another battery of emf 4V and internal resistance `1(Omega)`is joined to the point A as shown in figure.Take the potential at B to be zero. (a) What are the potentials at the poits A and C ? (b) At which point D of the wire AB, the potential is equal to the potential at C?(c ) If the point C and D are connected by a wire, what will be the current through it ? (d) If the 4V battery is replaced by 7.5V battery,wht would be the answer of parts (a) and (b) ?

Explanation

Potential difference between AB is 6V. B
is at 0 potential. Thus potential of a point
is 6V.
The potential difference between AC is 4V.
`V_A- V_C = 4 `
` V_C = V_A -4 = 6-4 = 2V. `
(b) The potential difference at D =` 2V = V_AD = 4V`.
` V_BD = 0V `
Current through the resisters R_1 & R_2 are
equal. Thus ,
` 4/R_1 = 2/R_2 `
` rArr R_1/R_2 = 2 `
` rArr l_1/l_2 = 2 `
(Acc.to the law of potentiometer)
` rArr l_1 + l_2 = 100 cm `
` rArr l_1 + (l_1/2) = 100 cm `
` rArr 3l_1/2 = 100cm `
` rArr = l_1 = 200/3 cm = 66.67 cm .`
` AD = 66.67 cm`
(c)When the points C and D are connected
by a wire, current flowing through it is 0,
since the points are equipotential .
(d) Potential at A = 6V.
` Potential at C = 6 - 7.5 =- 1.5 V. `
The potential at B= 0 and towards A
potential increases. Thus negative
potential point does not come within the
wire.