Physics Current Electricity

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

Initially,  $\frac{P}{Q}=\frac{40}{60}=\frac{2}{3}$ - (1)

Finally,  $\frac{P+x}{Q}=\frac{80}{20}=\frac{4}{1}$

(2) (1)

$\frac{P+x}{P}=\frac{4*3}{2}=6$

$\Rightarrow 1+\frac{x}{P}=6$

$\frac{x}{P}=5$

x = 5 P = 5 * 4 = 20 $\Omega$

$R=\frac{8}{2}\Omega =4\Omega$  (wheat stone bridge)

$I=\frac{V}{R}=\frac{40}{4}=10A$

$Current in circuit i=\frac{10}{4+1}=2 A$

$Terminal voltage =E-iR$

$=10-2*1=8 V$

$i=\frac{\epsilon }{R+r}$

$v_R=\frac{\epsilon R}{R+r}$

$\frac{1.25=\frac{\epsilon \left(5\right)}{5+r}}{1=\frac{\epsilon \left(2\right)}{2+r}}$

r = 1

Using above equ : $\epsilon =\frac{3}{2}=\frac{15}{10}V$

$R_1={?}_1\frac{\mathcal{l}}{A}{R}_2={?}_2\frac{\mathcal{l}}{A}$

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{A}{\mathcal{l}}\left[\frac{1}{{?}_1}+\frac{1}{{?}_2}\right]$

$=\frac{3*10^{?6}}{\frac{1}{4}}\left[\frac{1}{1.7*10^{?8}}+\frac{1}{2.6*10^{?8}}\right]$

$R=0.858*10^{?3}?$

Here

$R_{eq}=\left(12/{/}^{\mathcal{l}}6\right)+\left(4/{/}^{\mathcal{l}}4\right)+\left(6/{/}^{\mathcal{l}}12\right)$

$=\left(\frac{12*6}{18}\right)+\left(\frac{4*4}{8}\right)+\left(\frac{6*12}{18}\right)$

= 4 +2 + 4 = 10 $\Omega$

$l=JAcos\theta \Rightarrow 5=J*0.04*cos60°\Rightarrow J=\frac{5}{0.02}=250A/m^2$

$J=\frac{E}{\rho }\Rightarrow E=\rho J=44*10^{-8}*250=11*10^{-5}V/m$              

$l_1=\frac{50}{2000}=25mA,andl=\frac{V_i-V_Z}{1000}=\frac{100-50}{1000}=50mA$

$\Rightarrow l_z=l-l_1=25mA$

Method-I : Using Kirchhoff's Law for loop $\overrightarrow{BDABB},$  we can write

$-140+20l+6l_1=0\Rightarrow 10l+3l_1=70.......\left(i\right)$              

Using Kirchhoff's Law for loop $\overrightarrow{ACBBA},$  

we can write

$5\left(l-l_1\right)+90-6l_1=0$              

->-5l + 11l₁ = 90 .(ii)

Adding equation (1) with twice the equation (2), we have

$25l_1=250\Rightarrow l_1=\frac{250}{25}=10A$              

Method-II : Using concept of equivalent cell, we can write

$V_{AB}=\frac{\frac{140}{20}+\frac{0}{6}+\frac{90}{5}}{\frac{1}{20}+\frac{1}{6}+\frac{1}{5}}=\frac{2100+0+5400}{15+50+60}$

$=\frac{7500}{125}=60Volt$

$\Rightarrow l_1=\frac{60}{6}=10A$