Question

what is the value of shunt resistance that allows 20% of the main current through a galvanometer of 99 Ω​

Answer 1

$\large \bf \clubs \: To \: Find :-$

THE VALUE OF STUNT RESISTENCE OF 99

24.75 ohms.

$\underline{\sf{\bigstar\:According\:to\:the\:given\:Question\::-}}$

One way to handle parallel circuits is to convert to conductance.

The conductance of the galvanometer of 99 ohms is 1/99 = 0.010101

The conductance of the added resistor needs to be 4 times the galvanometer to have 20% go through the galvanometer.

Added resistor conductance is 4 * 0.0101010 = 0.040404

1/ 0.040404 = 24.75 ohms

Answer 2

Given: 20% of the main current is allowed through a galvanometer of 99 Ω​.

To find: The value of shunt resistance.

Solution:

• According to the formula,

        $S = (\frac{I_{g}}{I - I_{g}} ) G$

• Here, S is the shunt resistance, I is the main currect, $I_{g}$ is the current which on passing, produces full-scale deflection and G is the resistance in the galvanometer.
• According to the question, the galvanometer current is 20% os the main current.

• Hence, $I_{g} = \frac{20}{100} * I$

                         $= 0.2 I$

• The galvanometer resistance is given as 99Ω.
• On subsituting, we have,

        $S = (\frac{0.2 I}{I - 0.2 I} ) * 99$

           $= 24.75$Ω

Therefore, the the value of shunt resistance that allows 20% of the main current through a galvanometer of 99 Ω​ is 24.75Ω.