For loop BCD, equivalent resistance
\({ R }_{ 1 }=5\Omega +5\Omega =10\Omega \)
Across BA, equivalent resistance R₂ is
\(\cfrac { 1 }{ { R }_{ 2 } } =\cfrac { 1 }{ 15 } +\cfrac { 1 }{ 30 } +\cfrac { 1 }{ 10 } =\cfrac { 1 }{ 5 } \)
\({ R }_{ 2 }=5\Omega \)
\(\therefore\)  Potential difference V_AB = I x R₂ = 0.2 x 5
V_AB = 1V = -1V

(i) Applying Kirchhoff's second rule (law) in the closed loop ABFEA 
= 0.5 x 2 = VA - VB - 3
= 1 + 3 = VA - VB
VA - VB = 2V
The potential drop across R is 2V as R, EF & AB are parallel.
Applying Kirchoff's first rule at E
\(\Rightarrow\) 0.5 + I₂ = I
Where 'I' flows through 'R'
(ii) Now, Kirchhoff's 2^nd rule in closed loop FEABF
= 2I₂ + 0.5 x 2 = -4 + 3
= 2I₂ + 1 = -1
2I₂ = -2
The current in arm EF = I₂ = + 1A.

For the bridge to be balanced \(\cfrac { \\ P }{ Q } =\cfrac { R }{ S } \)
Since additionally a resistance 'x' is added in series with S, equation (1) can be written as
\(\cfrac { \\ P }{ Q } =\cfrac { R }{ (S+x) } \)
\(\left( S+x \right) =\cfrac { QR }{ P } x=\cfrac { \theta R }{ P } -S\)
\(x=\cfrac { 8\times 12 }{ 7 } -7\)
\(x=\frac{96}{7}-7=6.714 \Omega\)

Diameter d 0.24 cm = 0.24 x 10^-2 m
radius \(r=\cfrac { d }{ 2 } =0.12\times { 1 }^{ -2 }m\)
Current, I = 10A
Current density \(J=\cfrac { 1 }{ A } =\cfrac { 1 }{ { \pi r }^{ 2 } } \)
= \(\cfrac { 10 }{ 3.14\times \left( 0.12\times { 10 }^{ -2 } \right) ^{ 2 } } \)
= 2.2 x 106 Am^-2

According to Wheatstone bridge \(\cfrac { P }{ Q } =\cfrac { R }{ S } \)
\(\cfrac { R }{ S } =\cfrac { OA }{ OB } =\cfrac { 40 }{ 60 } \)
\(\cfrac { R }{ S } =\cfrac { OA }{ OB } =\cfrac { 40 }{ 60 } \)
If resistance 10 \(\Omega \) connected in series with R, the balance length is 60 cm.
\(\cfrac { R+10 }{ S } =\cfrac { 60 }{ 40 } \Rightarrow 2R+20=3S\)
From (1) & (2) \(\Rightarrow \left[ \cfrac { 4S }{ 3 } +20=3S \right] \)
45 + 60 = 95
55 = 60
\(S=\cfrac { 60 }{ 5 } =12\)
\(2\times \cfrac { 2S }{ 3 } +20=3S\)
\(s=12\Omega \)
From equation (1)
\(R=2\times \cfrac { 12 }{ 3 } =8\Omega \)
\(R=8\Omega \)