Question

Figure shows a Wheatstone bridge circuit. Which of the following correctly shows the currents $I_1,I_2$ and $I_3$ in the decreasing order of magnitude ?

• A. $I_1,I_2,I_3$
• B. $I_3,I_2,I_1$
• C. $I_2,I_3,I_1$
• D. $I_2,I_1,I_3$

Answer 1

The correct option is D $I_2,I_1,I_3$

Given Wheatstone bridge is balanced, because ratio of resistance side is equal i.e.,

$\frac{100\Omega }{50\Omega }=\frac{100\Omega }{50\Omega }$

Since bridge is balanced so there will not be current flow through the $25\Omega$.

$\therefore I_3=0$

current $I_1$ has lower magnitude as compare to $I_2$ because it passes through higher resistance $\left(200\Omega \right)$.

So, $I_2>I_1>I_3$

Hence, option $\left(c\right)$ is correct.