On bridges

We first let the unknown values of the resistances as $x$ and $y$ .

Then, we note the relationship of resistors in a balanced Wheatstone bridge (we regard it as balanced when $V_O = 0$ . That is,

$R_1 \cdot R_3 = R_2 \cdot R_4$

Simplifying, we get

$xy = 30240$ .

This will be our first equation.

Now, by means of voltage divider theorem, and letting $z$ be the total resistance of the bridge network, we can derive the following equation.

$\frac {1955}{2743} = 5\cdot \frac{ \frac {391}{13} } { \frac{391}{13} + z}$

$\frac {1955}{2743} = \frac {1955}{391 + 13z}$

It will be easy to determine that $z = \frac{2352}{13}$ .

Now, we get the values of $x$ and $y$ by simplifying the resistance network, and equating it to $z$ . That is,

$\frac { (x + 252)(y + 120) }{ x + y + 372 } = \frac { 2352} {13 }$

Simplifying that, we get the equation $x - \frac{7}{6}y + 112 = 0$

Then, we integrate that with the first found equation, so we get the values

$x = 140\: \Omega$ $y = 216\: \Omega$

So the answer is $x + y = \boxed { 356 \: \Omega }$ . Please don't forget to like/reshare!