Question # In the circuit shown in the figure.

# In the circuit shown in the figure.

[

E_{1}=10 mathrm{V}, E_{2}=4 mathrm{V}, r_{1}=r_{2}=1 Omega text { and } R=2 Omega

]

Find the respective potential difference across battery 1 and battery 2

( A quad 5.5 vee ) each

B ( 8.5 mathrm{V} ) each

( mathrm{C} quad 5.5 mathrm{V}, 8.5 mathrm{V} )

D ( 8.5 vee, 5.5 vee )

Solution

[

begin{array}{l}

E_{n c t}=10-4=6 V

r_{n c t}=r_{1}+r_{2}+R=1+1+2=4 Omega

i=frac{E_{n c t}}{r_{n e f}}=frac{6}{4}=frac{3}{2} A

end{array}

]

Poteutial difference across bathery ( 1=E_{1}-operatorname{lr} )

[

=10-frac{3}{2}(1)=frac{17}{2}=8 cdot 5 mathrm{V}

]

Poloutial difference acros battery ( 2=E_{2}+i r_{2} )

[

=4+frac{3}{2}(1)=frac{11}{2}=5 cdot 5 v

]

option (D)