V = 12 V and R = 24 Ω
Current, I = ?
From Ohm’s law, \(I=\frac{V}{R}=\frac{12}{24}=0.5A\)

R_s = R₁ + R₂ = 15 Ω           (1)
\({ R }_{ p }=\frac { { R }_{ 1 }{ R }_{ 2 } }{ { R }_{ 1 }{ +R }_{ 2 } } =\frac { 56 }{ 15 } \Omega \quad \) (2)
From equation (1) substituting for R₁ + R₂ in equation (2)
\(\frac { { R }_{ 1 }{ R }_{ 2 } }{ 15 } =\frac { 56 }{ 15 } \Omega \)
∴ R₁R₂ = 56
\({ R }_{ 2 }=\frac { 56 }{ 15 } \Omega \)         (3)
Substituting for R₂ in equation (1) from equation (3)
\({ R }_{ 1 }+\frac { 56 }{ { R }_{ 1 } } =15\)
Then, \(\frac { { R }_{ 1 }^{ 2 }+56 }{ { R }_{ 1 } } =15\)
R₁² + 56 = 15 R₁
R₁² - 15 R₁ + 56 = 0
The above equation can be solved using factorisation.
R₁ = 8 Ω (or) R₁ = 7 Ω
If (R₁ = 8 Ω)
Substituting in equation (1)
8 + R₂ = 15
R₂ = 15 – 8 = 7 Ω ,
R₂ = 7 Ω i.e , (when R₁ = 8 Ω ; R₂ = 7 Ω)
If R₁= 7 Ω
Substituting in equation (1)
7 + R₂ = 15
R₂ = 8 Ω , i.e , (when R₁ = 7 Ω ; R₂ = 8 Ω )

Since the resistances are connected in parallel, therefore, the equivalent resistance in the circuit is
\(\frac { 1 }{ { R }_{ p } } =\frac { 1 }{ { R }_{ 1 } } +\frac { 1 }{ { R }_{ 2 } } =\frac { 1 }{ 4 } +\frac { 1 }{ 6 } \)
\(\frac { 1 }{ { R }_{ p } } =\frac { 5 }{ 12 } \Omega \quad or\quad { R }_{ p }=\frac { 12 }{ 5 } \Omega \)
The resistors are connected in parallel, the potential difference (voltage) across them is the same.
\({ I }_{ 1 }=\frac { V }{ { R }_{ 1 } } =\frac { 24V }{ 4\Omega } =6A\)
\({ I }_{ 2 }=\frac { V }{ { R }_{ 2 } } =\frac { 24 }{ 6 } =4A\)
The current I is the sum of the currents in the two branches. Then,
I = I₁ + I₂ = 6 A + 4 A = 10 A

Since the resistors are connected in series, the effective resistance in the circuit
= 4 Ω + 6 Ω = 10 Ω
The Current I in the circuit =\(\frac { V }{ { R }_{ eq } } =\frac { 24 }{ 10 } =2.4A\)
Voltage across 4Ω resistor
V₁= IR₁ = 2.4A x 4Ω = 9.6V
Voltage across 6 Ω resistor
V₂ = IR₂ = 2.4A x 6Ω = 14.4V

In the first case, the resistance of the block
\({ R }_{ AB }=\rho \frac { length }{ Area } =\rho \frac { C }{ AB } \)
The current \({ I }_{ AB }=\frac { V }{ { R }_{ AB } } =\frac { V }{ \rho } .\frac { AB }{ C } \quad (1)\)
In the second case, the resistance of the block \({ R }_{ BC }=\rho \frac { A }{ BC } \)
The current \({ I }_{ BC }=\frac { V }{ { R }_{ BC } } =\frac { V }{ \rho } .\frac { BC }{ C } \quad (2)\)
To express I_BC interms of I_AB, we multiply and divide equation (2) by AC, we get
\({ I }_{ BC }=\frac { V }{ \rho } .\frac { BC }{ A } \frac { AC }{ AC } =\left( \frac { V }{ \rho } .\frac { AB }{ C } \right) .\frac { { C }^{ 2 } }{ { A }^{ 2 } } =\frac { { C }^{ 2 } }{ { A }^{ 2 } }.{ I }_{ AB }\)
Since C > A, the current I_BC > I_AB

R₁ = 20 Ω, R₂ = ?
Let the original length of the wire (l₁) be l.
New length, l₂ = 8l₁ (i.,e) l₂ = 8l
Original resistance, R₁ = \(\rho \frac { { l }_{ 1 } }{ { A }_{ 1 } } \)
New resistance R₂ = \(\rho \frac { { l }_{ 2 } }{ { A }_{ 2 } } =\frac { \rho (8l) }{ { A }_{ 2 } } \)
Though the wire is stretched, its volume is unchanged.
Initial volume = Final volume
A₁l₁ =  A₂l₂ , A₁l = A₂(8l)
\(\frac { { A }_{ 1 } }{ { A }_{ 2 } } =\frac { 8l }{ l } =8\)
By dividing equation R₂ by equation R₁, we get
\(\frac { { R }_{ 2 } }{ { R }_{ 1 } } =\frac { \rho (8l) }{ { A }_{ 2 } } \times \frac { { A }_{ 1 } }{ \rho l } \)
\(\frac { { R }_{ 2 } }{ { R }_{ 1 } } =\frac { { A }_{ 1 } }{ { A }_{ 2 } } \times 8\)
Substituting the value of \(\frac { { A }_{ 1 } }{ { A }_{ 2 } } \), we get
\(\frac { { R }_{ 2 } }{ { R }_{ 1 } } =8\times 8=64\)
R₂ = 64 x 20 = 1280 Ω
Hence, stretching the length of the wire has increased its resistance.