Question

A circuit is shown in diagram. If we put a conducting wire AB as shown doted then current through wire will be

• A. Zero
• B. 1 amp
• C. -1 amp
• D. 3 amp

Answer 1

The correct option is D 3 amp

The equivalent circuit diagram of given circuit is shown in IMAGE 02

Here, $R_1$ and $R_2$ are in parallel,

So,

$R_{eq1}=\frac{R_1\cdot R_2}{R_1{+R}_2}=\frac{30\times 60}{30+60}=\frac{1800}{90}=20\Omega$

And,$R_3$ and $R_4$ are also in parallel.

$R_{eq2}=\frac{R_3\cdot R_4}{R_3{+R}_4}=\frac{60\times 30}{60+30}=\frac{1800}{90}=20\Omega$

Now, $R_{eq1}$ and $R_{eq2}$ are in series,

$R_{eq}=R_{eq1}+R_{eq2}$

$\Rightarrow R_{eq}=20+20=40\Omega$

so, the current $\left(I\right)=\frac{V}{R}=\frac{120V}{40\Omega }=3Amp$