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Answer 1

Answer:

Suppose a current i_(1) goes in the branch BAD and current i_(2) in the branch DCB. The current in DB will be i_(1)-i_(2) from the junction law. The circuit with the currents shown is redrawn in figure. Applying the loop law to BADB we ge,

(2(Ω))i1−2V+1V+(1(Ω))i1+(2(Ω))(i1−i2)=0

(5(Ω))i1−(2(Ω))i2=1V....(i)

Applying the same law to the loop DCBD,we get

−3V+(3(Ω)i2+(1(Ω))i2+1V−(2(Ω))(i1−i2)=0

or,−(2(Ω))i1+(6(Ω))i2=2V....(ii)

From (i)and (ii),

i1=513A,i2=613A.

sotˆi_(1)-i_(2)=1/13A.thecurrent∈BDisomB→D.(a)V_(B)-V_(D)=(2(Omega)((1)/(13)A)=(2)/(13)V.The potential difference across the cell G is

Vc−Vd=−(3(Ω))i2=3V.

=(3V−1813V)=2113V.

The potential d⇔erenceacrossthecellHisV_(C)-V_(B)=(1(Omega))i_(2)+1V=(1(Omega))((6)/(13)A+1V=(19)/(13)V.`