Electric Current

In first condition R₁ = 36 $\Omega$  

In second condition R₂ = 18  $\Omega$

$P_1=\frac{V^2}{R_1}=\frac{{\left(240\right)}^2}{36}$               

$P_2=\frac{V^2}{R_2}+\frac{V^2}{R_2}=\frac{{\left(240\right)}^2}{18}+\frac{{\left(240\right)}^2}{18}$               

$P_2=\frac{{\left(240\right)}^2}{9}$               

So   $\frac{P_1}{P_2}=\frac{{\left(240\right)}^2/36}{{\left(240\right)}^2/9}=\frac{1}{4}$

x = 4

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{4}+\frac{1}{4}$  

 R_eq = 2

$\frac{\Delta R_{eq}}{R_{eq}^2}=\frac{\Delta R_1}{R_1^2}+\frac{\Delta R_2}{R_2^2}$

$\begin{array}{l}\frac{\Delta R_{eq}}{4}=\frac{0.8}{16}+\frac{0.4}{16}+\frac{1.2}{16}\\ R_{eq}=0.3\end{array}$                

In given circuit inductor behave as a simple wire so resultant circuit will be

R_ef = 2 + 1 = 3 $\Omega$  

V = IR

$l=\frac{30}{3}=10A$  

Energy required to melt

Q = $MS\Delta T+ML$  

$\Rightarrow 10^{-1}*2*10^3*10+10^{-1}*3.33*10^5$      

->3.53 * 10⁴ J

Heat produce in wire

H = l²RT

  $Q=3.53*10^4={\left(\frac{1}{2}\right)}^2*\left(4*10^3\right)*t$

$t=\frac{3.53*10^4*4}{4*10^3}=35.3sec$             

When switch is open,

$R_{eq}=\frac{3R}{2}$              

When switch is closed,

$R_{eq}^{\text{'}}=2*\frac{R*2R}{R+2R}=\frac{4R}{3}$           

$\frac{R_{eq}}{R_{eq}^{\text{'}}}=\frac{3R/2}{4R/3}=\frac{9}{8}=\frac{x}{8}$

$\Rightarrow x=9$

Current at full deflection,

$l_{max}=\frac{50}{2}=25mA$             

$R=\frac{V}{l_{max}}=\frac{50}{25}=2K\Omega$                

$R=75*10^2\pm 5\mathrm{\%}$

$\Rightarrow R=7500\pm 375\Omega$