Question 270: Current Electricity

Question

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 :

Accepted Answer

Equivalent resistance of circuit
$$
\begin{aligned}
\mathrm{R}_{\mathrm{eq}} & =3+1+2+4+2 \\
& =12 \Omega
\end{aligned}
$$
Current through battery $\mathrm{i}=\frac{24}{12}=2 \mathrm{~A}$
$$
\begin{aligned}
& \mathrm{I}_4=\frac{\mathrm{R}_5}{\mathrm{R}_4+\mathrm{R}_5} 	imes 2=\frac{5}{20+5} 	imes 2=\frac{2}{5} \mathrm{~A} \\
& \mathrm{I}_5=2-\frac{2}{5}=\frac{8}{5} \mathrm{~A}
\end{aligned}
$$

Answer

$$\mathrm{I_4=\frac{8}{5}A}$$ and $$\mathrm{I_5=\frac{2}{5}A}$$

Answer

$$\mathrm{I_4=\frac{6}{5}A}$$ and $$\mathrm{I_5=\frac{24}{5}A}$$

Answer

$$\mathrm{I_4=\frac{2}{5}A}$$ and $$\mathrm{I_5=\frac{8}{5}A}$$

Answer

$$\mathrm{I_4=\frac{24}{5}A}$$ and $$\mathrm{I_5=\frac{6}{5}A}$$

Current Electricity

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 :

Choose the correct answer: