Question

A 6-volt battery of negligible internal resistance is connected across a uniform wire AB of length 100 cm. The positive terminal of another battery of emf 4 V and internal resistance 1 $\Omega$ is joined to the point A as shown in figure (32-E28). Take the potential at B to be zero. (a) What are the potentials at the points A and C ? (b) At which point D of the wire AB, the potential is equal to the potential at C ? (c) If the points C and D are connected by a wire, what, will be the current through it ? (d) If the 4 V battery is replaced by 7.5 V battery, what would be the answers of parts (a) and (b) ?

figure (32-E28)

Answer 1

Potential difference between AB is 6 V. B is at 0 potential. Thus potential of a point is 6 V.

The potential difference between AC is 4V.

$V_A-V_C=4$

$V_C=V_A-4=6-4=2V$

(b) The potential at $D=2V=V_{AD}=4V.$ $V_{BD}=0V.$

Current through the resisters $R_1$ & $R_2$ are equal. Thus

$\frac{4}{R_1}=\frac{2}{R_2}$

$\Rightarrow \frac{R_1}{R_2}=2$

$\Rightarrow \frac{l_1}{l_2}=2$

(Acc. to the law of potentiometer)

$\Rightarrow l_1+l_2=100\text{cm}$

$\Rightarrow l_1+\frac{l_1}{2}100\text{cm}$

$\Rightarrow \frac{3l_1}{2}=100\text{cm}$

$\Rightarrow l_1=\frac{200}{3}\text{cm}=66.67\text{cm}$

$\text{AD}=66.67\text{cm}$

(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.