Question 593: Current Electricity

Question

Consider the circuit shown in the figure. The value of current 'I' is

Accepted Answer

Applying Kirchoff's voltage law to loop ABEFA we get
$$\begin{aligned}
& -28 \mathrm{I}_1-6-8=0 \\
& -28 \mathrm{I}_1-14=0 \\
	herefore & -28 \mathrm{I}_1=14 \\
	herefore \quad & \mathrm{I}_1=\frac{-14}{28}=-\frac{1}{2} \mathrm{~A}
\end{aligned}$$
Applying KVL to BCDEB
$$\begin{aligned}
& 54 \mathrm{I}_2-12+6=0 \\
	herefore \quad & 54 \mathrm{I}_2=6 \\
	herefore & \mathrm{I_2=\frac{6}{54}=\frac{1}{9} A} \quad 	herefore \mathrm{I=I_1+I_2=-\frac{1}{2}+\frac{1}{9}=-\frac{7}{18} A}\end{aligned}$$

Answer

$$-\frac{7}{18}$$A

Answer

5A

Answer

3A

Answer

$$-3$$A

Current Electricity

Consider the circuit shown in the figure. The value of current 'I' is

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