Ohms Law

Let V  be the potential of the junction as shown in figure. Applying junction law, we have

or $\frac{20-V}{2}+\frac{5-V}{4}\overline{\to }\frac{V-0}{2}$

or $40-2V+5-V=2V$ or $5V=45\Rightarrow V=9V$

$\therefore i_3=\frac{V}{2}=4.5A$

$\begin{array}{l}{D}_1\\ D_3\end{array}\}$ Forward biased offer zero Resistance

D₂} Reversed biased offers Infinite Resistance

$I=\frac{v}{R}=\frac{10}{10}=1Amp$

$\equiv Req=2.5\Omega$

$l=\frac{v}{R_{eq}}=\frac{5}{2.5}=2A$

R_eq = 10 + (50 * 20) / (50 + 20) = 170/7 Ω
⇒ I = 170 / (170/7) = 7A
⇒ x = 10 * 7 = 70V ⇒ Voltage across 10Ω resistor

From ohm's Law, V = IR = I (ρl / A) = I (ρl / (πd²/4) ⇒ ρ = (πd²V) / (4lI)
Relative error in resistivity,
Δρ/ρ = 2 (Δd/d) + ΔV/V + Δl/l + ΔI/I = 2 * (0.01/5.00) + (0.1/5.0) + (0.1/10.0) + (0.01/2000) = 0.039
Percentage error = (Δρ/ρ) * 100 = 3.9%