In the circuit shown in the given figure the resistance $R_1$ and $R_2$ are respectively
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$\mathrm{V}_{\mathrm{BE}}=\mathrm{V}_{\mathrm{CD}}=\mathrm{V}_{\mathrm{CE}}$

$\Rightarrow 0.5 \mathrm{R}_{2}=1 \times 20$

$\Rightarrow \mathrm{R}_{2}=40\, \Omega$

$10 \mathrm{i}=1 \times 20 \Rightarrow \mathrm{i}=2 \mathrm{\,A}.$

$\therefore \mathrm{i}_{0}=0.5+1+2=3.5 \mathrm{\,A}$

$\therefore \mathrm{V}_{\mathrm{AB}}=69-\mathrm{V}_{\mathrm{BE}}$

$\Rightarrow \mathrm{i}_{0} \mathrm{R}_{1}=69-20$

$\Rightarrow \mathrm{R}_{1}=14\, \Omega$

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