Science Answer the questions.[Phy-2M]

$\therefore \ \text{I}=\frac{96000}{3600}=\frac{80}{3} \ \text{A}$

$\text{H}=50\times\frac{80}{3}\times60\times60=4.8\times10^6 \ \text{J}$

$\frac{1}{\text{Req}}=\frac{1}{\text{R}}+\frac{1}{\text{R}}+\frac{1}{\text{R}}........\text{N time}$

$\frac{1}{\text{Req}}=\frac{\text{N}}{\text{R}}$

$\therefore\bigg(\text{Req}\bigg)_2=\frac {\text{R}}{\text{N}}$

$\therefore\frac{(\text{Req})}{(\text{Req})_2}=\frac{\text{NR}\times\text{N}}{\text{R}}=\text{N}^2$

So, Rs₁ = 4 + 8 = 12

Similary R₂ and R₃ are in series

Rs₂ = Q + 4 = 12

Now Rs₁ and Rs₂ are in || combination

So, $\frac{1}{\text{RP}}=\frac{1}{\text{12}}+\frac{1}{\text{12}}=\frac{1}{\text{12}}=\frac{1}{\text{6}}$

RP = 6

So, R_eq = $6\Omega$

$\text{W}_\text{B}= -\frac{2\text{C}}{5}\text{C}-\frac{1}{5}\text{C}=-\frac{3}{5}\text{C}$

$\text{W} = -\frac{2}{5}\text{C} - \frac{1}{5}\text{C} = -\frac{3}{5}\text{C}$

$\text{W} = \frac{\text{q}}{\text{v}}$

$\text{WA} = \frac{2\text{C}}{10} = \frac{1}{5}\text{C}$

$\frac{1}{\text{R}}=\frac{1}{\text{R}1}+\frac{1}{\text{R}1}$

$=\frac{1}{10}+\frac{1}{10}$

$=\frac {2}{10} \text{ohms,}$

$\text{R} = 5 \text{ohms.}$

$\text{R}=\frac{120}{2.4}=50\ \text{ohm}$

$\therefore\text{H}=(15)^{2}\times500\times1800\text{J}$

$\text{H}=225\times500\times1800\text{J}$

$\therefore\text{H}=20.25\times10^{7}\text{J}$

$\frac{100}{4}=\text{I}^2$