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 4V and internal resistance `1 Omega` is joind to the point A as shown in figure. Take the potential at B to be zero. (a) What are the potentials at the points A and C ? (b) At which 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 is ? (d) Id the 4V batteryt is replaced by 7.5 V battery, what whould be the answers of parts (a) and (b) ?

Answer

A, B, C, D

Explanation

a. `V_A-V_B=6V`
Siince `V_B=0`
`:. V_A=6V`
`V_A-V_C=4VimpliesV_C=V_A-4=2V`
b. `V_A-V_D=V_A-V_C=4V`
From unitary method we can find that
`AD=(100/6)(4)=66.67cm`
c. Since, they are at same potetial no current will flow through it
d. ` V_A-V_B` is still `6V :. V_A=6V`
Further `V_A-V_C=7.5V`
`:. V_C=-1.5V`
Since `EMF` of the battery in lower circuit is more than `EMF` of the battery in upper circuit. No such point will exist.