Question

The resistance in four arms of a Wheatstone's bridge are \[3\Omega ,\,6\Omega ,{\text{ }}9\Omega {\text{ }}and{\text{ }}30\Omega \] respectively. What shunt is needed across \[30\Omega \] resistance to balance the network?

Answer 1

Hint: In order to answer this question, first we will write the effective resistance, in parallel across \[30\Omega \] . And then we will follow the formula of shunt resistance in terms of all 4 given resistance of a Wheatstone’s Bridge.

Complete step by step solution:
The Wheatstone Bridge Circuit consists of a bridge formed by two known resistors, one unknown resistor, and one variable resistor. This bridge is extremely dependable since it provides precise measurements.
If the bridge is to be balanced then-
If a resistance is $R$ but in parallel across \[30\Omega \] , the effective resistance in the fourth arm will be $\dfrac{{30R}}{{30 + R}}$ .
So, the required shunt resistance will be:
$
  \therefore \dfrac{3}{6} = \dfrac{9}{{\dfrac{{30R}}{{30 + R}}}} \\
   \Rightarrow \dfrac{1}{2} = \dfrac{{9(30 + R)}}{{30R}} \\
   \Rightarrow \dfrac{1}{2} = \dfrac{{3(30 + R)}}{{10R}} \\
   \Rightarrow 6(30 + R) = 10R \\
   \Rightarrow 180 + 6R = 10R \\
   \Rightarrow 10R - 6R = 180 \\
   \Rightarrow 4R = 180 \\
   \Rightarrow R = 45\Omega \\
 $
Hence, the required shunt resistance is $45\Omega $ .

Additional Information:- Shunt resistance refers to a resistor with a very low resistance value. The main component of a shunt resistor is a low-temperature coefficient of resistance material. It's wired in series with the ammeter whose range needs to be increased.

Note: By offering an additional current path for the light-generated current, low shunt resistance causes power losses in solar cells. The amount of current going through the solar cell junction is reduced, and the voltage from the solar cell is reduced as a result of this diversion.