Q = 15 Ω, l₁: l₂ = 3 : 2
\(\frac { { l }_{ 1 } }{ { l }_{ 2 } } =\frac { 3 }{ 2 } \)
\(\frac { P }{ Q } =\frac { { l }_{ 1 } }{ { l }_{ 2 } } \)
\(P=Q\frac { { l }_{ 1 } }{ { l }_{ 2 } } \)
\(P=15 \times \frac { 3 }{ 2 } =22.5\Omega \)

The given values I = 3.93 A, ξ = 12 V,
R = 3 Ω
(a) The terminal voltage of the battery is equal to voltage drop across the resistor
V = IR = 3.93 x 3 = 11.79 V
The internal resistance of the battery,
\(r=\left[ \frac { \xi -V }{ V } \right] R=\left[ \frac { 12-11.79 }{ 11.79 } \right] \times 3=0.05\Omega \)
(b) The power delivered by the battery P = Iξ = 3.93 x 12 = 47.1 W
The power delivered to the resistor = I² R = 46.3 W
The remaining power = (47.1 – 46.3) = 0.8 W is delivered to the internal resistance and cannot be used to do useful work. (it is equal to I² r).

\(\frac { P }{ Q } =\frac { R }{ S } \)
\(S=\frac { Q }{ P } \times R\)
\(S=\frac { 1000 }{ 100 } \times 40S=400\Omega \)

Thus applying Kirchoff’s second law to the closed loop EACE
I₁R₁ + I₂R₂ + I₃R₃ = ξ
and for the closed loop ABCA
I₄R₄ + I₅R₅ - I₂R₂ = 0

i) Equivalent emf ξ_eq = 5 V
ii) Equivalent internal resistance,
\({ R }_{ eq }=\frac { r }{ n } =\frac { 0.5 }{ 4 } =0.125\Omega \)
iii) total current, \(I=\frac { \xi }{ { R }_{ 5 }+\frac { r }{ n } } \)
\(I=\frac { 5 }{ 10+0.125 } =\frac { 5 }{ 10.125 } \)
I ≈ 0.5 A 
iv) Potential difference across each cell
V = IR = 0.5 x 10 = 5 V
v) Current from each cell, \(I'=\frac { I }{ n } \)
\(I'=\frac { 0.5 }{ 4 } =0.125A\)