$\frac{2}{x}=\frac{\ell+0.2}{1-(\ell-0.2)}=\frac{\ell+0.2}{0.8-\ell}$
$\frac{\ell(\ell+0.2)}{(1-\ell)(0.8-\ell)}=1$
$\ell^{2}+0.2 \ell=0.8+\ell^{2}-0.8 \,\ell-\ell$
$2\, \ell=0.8$
$\ell=0.4 \mathrm{\,m}$
$\frac{x}{2}=\frac{0.4}{0.6} \Rightarrow x=\frac{4}{3}$
$\frac{2}{x}=\frac{\ell-0.2}{1-(\ell-0.2)}=\frac{\ell-0.2}{1.2-\ell}$
$1=\frac{\ell(\ell-0.2)}{(1-\ell)(1.2-\ell)}$
$\ell-0.2\, \ell=1.2+\ell^{2}-2.2\, \ell$
$2 \ell=1.2$
$\ell=0.6$
$\Rightarrow x=\frac{2 \times 0.6}{0.4}=3\, \Omega$
($A$) The voltmeter displays $-5 \mathrm{~V}$ as soon as the key is pressed, znd displays $+5 \mathrm{~V}$ after a long time
($B$) The voltmeter will display $0 \mathrm{~V}$ at time $t=\ln 2$ seconds
($C$) The current in the ammeter becomes $1 / e$ of the initial value after $1$ second
($D$) The current in the ammeter becomes zero after a long time
($1$) The magnitude of $q_1$ is
($2$) The magnitude of $q _2$ is
Give the answer of question ($1$) and ($2$)



The current in resistance $R _2$ would be zero if
$(A)$ $V_1=V_2$ and $R_1=R_2=R_3$
$(B)$ $V_1=V_2$ and $R_1=2 R_2=R_3$
$(C)$ $V_1=2 V_2$ and $2 R_1=2 R_2=R_3$
$(D)$ $2 V _1= V _2$ and $2 R _1= R _2= R _3$