Physics Current Electricity

According to question, we can write

$\frac{2V}{2+2r}=\frac{V}{2+\frac{r}{2}}$

$\Rightarrow 2+2r=4+r\Rightarrow r=2\Omega$  

$i=\frac{v_0}{R}\left(1-e^{\frac{tR}{L}}\right)$

$U=\frac{1}{2}L{i}^2$

$\frac{dU}{dt}=Li\frac{di}{dt}$

$2L*\frac{R}{L}*\frac{v_0^2}{R^2}\left(1-e^{\frac{tR}{L}}\right)e^{\frac{tR}{L}}=\frac{v_0^2}{R}\left(1-e^{\frac{tR}{L}}\right)$

$t=\frac{L}{R}In2$

$i=\frac{v}{R+G}$

$\frac{i}{2}=\frac{v}{\frac{SG}{S+G}+R}*\frac{S}{S+G}$

$\Rightarrow \frac{2S}{SG+RG+RS}=\frac{1}{R+G}$

$\frac{SR}{R-S}=G$

R > > S

$\Rightarrow G\approx S$

  $5*10^3{I}_1=60$

$\Rightarrow I_1=\frac{60}{5*10^3}=12mA$                  

Potential difference across

$10k\Omega =240-60=180V$            

$I_2=I_3-I_1=6mA$            

  $\boxed{I=\frac{E}{R_2}}$

$U_{Inductor}=Heatlossin{R}_1=\frac{1}{2}L{\left(\frac{E}{R_2}\right)}^2$              

Voltmeter readings

  $\begin{array}{ll}\Rightarrow & 20=I_{1}R\\ \Rightarrow & 20=\left(\frac{R_v}{R+R_v}\right)4R\\ \Rightarrow & R_{v}R=5R+5R_v\Rightarrow \mathrm{}R=\frac{5R_v}{R_v-5}\\ \therefore & R=\frac{5}{1-\frac{5}{R_v}}\end{array}$

For $5<{\mathrm{R}}_v<\mathrm{\infty },\mathrm{R}>5\mathrm{\Omega }$

$i_i=\frac{84}{140}=0.6A$

$i_2=\frac{84}{60}=1.4A$

p.d across capacitor = 20 v

Q = CV = 5 * 20 = 100    $?C$

After S is open

Q = Q . e^-t/RC

=100 e^-4

  $\frac{P}{Q}=\frac{R}{X}{X}_{max}=\frac{R_{max}.Q_{max}}{P_{min}}$

$=9000*\frac{1000}{10}=9*10^{5}ohm$

$x_{min}=\frac{R_{min}.Q_{min}}{P_{max}}=100*\frac{10}{1000}=1ohm$  

Gold band is equal to 5% tolerance