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 . Take the potential at B to be zero.  If  potential at (A) is $V_1$  and at (C) is  $V_2$   then $V_1+V_2$  =______

a) Potential difference between A and B is 6 V. 
B is at 0 potential.
Thus potential of A point is 6V.                                         
Potential difference between Ac is 4 V.
 $\begin{array}{l}{V}_A-V_C=4\\ V_C=V_A-4=6-4=2V.\end{array}$    

              
b) The Potential at  $D=2V,V_{AD}=4V;V_{BD}=0V$
Current through the resistors $R_{1}and{R}_2$  are equal.
Thus ,  $\frac{4}{R_1}=\frac{2}{R_2}$
 $\Rightarrow \frac{R_1}{R_2}=2$
$\Rightarrow \frac{I_1}{I_2}=2$  (Acc. to the law of potentiometer)
 $\begin{array}{l}{I}_1+I_2=100cm\\ \Rightarrow I_1+\frac{I_1}{2}=100cm\Rightarrow \frac{3I_1}{2}=100cm\\ \Rightarrow I_1=\frac{200}{3}cm=66.67cm.\\ AD=66.67cm\end{array}$
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=6 v
Potential at C = 6-7.5= 1.5 C
The potential at B = 0 and towards A potential increases.
Thus –ve potential point does not come within the wire.