Question 615: Current Electricity

Question

A constant voltage of 50 V is maintained between the points $A$ and $B$ of the circuit shown in the figure. The current through the branch $C D$ of the circuit is:

Accepted Answer

$$\begin{aligned}
& R_{A B}=(1 \Omega / 3 \Omega) 	ext { in series with }(2 \Omega / 4 \Omega) \
& =\frac{3 	imes 1}{3+1}+\frac{2 	imes 4}{2+4} \
& =\frac{3}{4}+\frac{8}{6}=\frac{9+16}{12}=\frac{25}{12} \Omega
\end{aligned}$$
Now total current through cell
$$\begin{aligned}
& I=\frac{50}{25 / 12}=24 \mathrm{~A} \
& I_{1 \Omega}=\frac{3}{4} 	imes 24=18 \mathrm{~A}, I_{3 \Omega}=\frac{1}{4} 	imes 24=6 \mathrm{~A} \
& I_{2 \Omega}=\frac{4}{6} 	imes 24=16 \mathrm{~A}, I_{4 \Omega}=\frac{2}{6} 	imes 24=8 \mathrm{~A}
\end{aligned}$$
Using junction rule at $C, I_{C D}=18-16=2 \mathrm{~A}$ (From $C$ to $D$)

Answer

2.5 A

Answer

3.0 A

Answer

1.5 A

Answer

2.0 A

Current Electricity

A constant voltage of 50 V is maintained between the points A and B of the circuit shown in the figure. The current through the branch C D of the circuit is:

Choose the correct answer: