Question

The current (i) in the network is

• A. $0A$
• B. $0.3A$
• C. $0.2A$
• D. $0.6A$

Answer 1

The correct option is B

$0.3A$

Step 1: Circuit diagram

First of all, rearrange the given figure, the diodes are reverse biased, they will not conduct, hence, the circuit will look like,

Step 2: Finding current

Now calculate$R_{eff}$.

Therefore,

$$\begin{gathered} R_{eff}=5+10+5+10 \\ {R}_{eff}=30\Omega \end{gathered}$$

Now calculate$I$, then

$$\begin{gathered} \Rightarrow I=\frac{V}{R_{eff}} \\ \Rightarrow I=\frac{9}{30} \\ \Rightarrow I=0.3A \end{gathered}$$

Therefore the current will be $0.3A$.

Hence, the correct option is (B).