The current flowing in a coil of resistance $90~\Omega$ is to be reduced by $90\%$. What value of resistance s

Question

The current flowing in a coil of resistance $90~\Omega$ is to be reduced by $90\%$. What value of resistance should be connected in parallel with it?

Accepted Answer

1. **Concept:** To reduce the current flowing through a coil (like a galvanometer), a low resistance called a **shunt** ($S$) is connected in parallel with it. This diverts a portion of the current .
2. **Data Given:**
   - Resistance of the coil, $G = 90~\Omega$.
   - The current in the coil ($I_g$) is to be reduced **by** $90\%$. This means the current remaining in the coil is $10\%$ of the main current ($I$).
   - Therefore, $I_g = 0.1 I$.
   - The remaining current passes through the shunt: $I_s = I - I_g = I - 0.1I = 0.9I$.
3. **Calculation:** Since the coil and the shunt are in parallel, the potential difference across them is equal.
   $$ V_g = V_s $$
   $$ I_g G = I_s S $$
   $$ (0.1 I) 	imes 90 = (0.9 I) 	imes S $$
   $$ 9 = 0.9 S $$
   $$ S = \frac{9}{0.9} = 10~\Omega $$
4. **Conclusion:** A resistance of $10~\Omega$ should be connected in parallel.

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