JEE MAIN - Physics (2023 - 24th January Morning Shift - No. 12)
As shown in the figure, a network of resistors is connected to a battery of 24V with an internal resistance of 3 $$\Omega$$. The currents through the resistors R$$_4$$ and R$$_5$$ are I$$_4$$ and I$$_5$$ respectively. The values of I$$_4$$ and I$$_5$$ are :
_24th_January_Morning_Shift_en_12_1.png)
$$\mathrm{I_4=\frac{8}{5}A}$$ and $$\mathrm{I_5=\frac{2}{5}A}$$
$$\mathrm{I_4=\frac{6}{5}A}$$ and $$\mathrm{I_5=\frac{24}{5}A}$$
$$\mathrm{I_4=\frac{2}{5}A}$$ and $$\mathrm{I_5=\frac{8}{5}A}$$
$$\mathrm{I_4=\frac{24}{5}A}$$ and $$\mathrm{I_5=\frac{6}{5}A}$$
Explanation
Equivalent resistance of circuit
$$ \begin{aligned} \mathrm{R}_{\mathrm{eq}} & =3+1+2+4+2 \\\\ & =12 \Omega \end{aligned} $$
Current through battery $\mathrm{i}=\frac{24}{12}=2 \mathrm{~A}$
$$ \begin{aligned} & \mathrm{I}_4=\frac{\mathrm{R}_5}{\mathrm{R}_4+\mathrm{R}_5} \times 2=\frac{5}{20+5} \times 2=\frac{2}{5} \mathrm{~A} \\\\ & \mathrm{I}_5=2-\frac{2}{5}=\frac{8}{5} \mathrm{~A} \end{aligned} $$
$$ \begin{aligned} \mathrm{R}_{\mathrm{eq}} & =3+1+2+4+2 \\\\ & =12 \Omega \end{aligned} $$
Current through battery $\mathrm{i}=\frac{24}{12}=2 \mathrm{~A}$
$$ \begin{aligned} & \mathrm{I}_4=\frac{\mathrm{R}_5}{\mathrm{R}_4+\mathrm{R}_5} \times 2=\frac{5}{20+5} \times 2=\frac{2}{5} \mathrm{~A} \\\\ & \mathrm{I}_5=2-\frac{2}{5}=\frac{8}{5} \mathrm{~A} \end{aligned} $$
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