The Question is from neet-physics , and it was created by the author VENKATESWARARAO

Question

A galvanometer with a resistance of $50 \Omega$ and a full-scale deflection current of $100 \mu A$ is to be converted into a voltmeter with a range of 0-5V.  A resistance $R_x$ is connected in series with the galvanometer.  Simultaneously, an identical galvanometer is converted into an ammeter of range 0-10mA by connecting a shunt resistance $R_s$ in parallel. What is the relationship between $R_x$ and $R_s$?

VENKATESWARARAO · Accepted Answer

## CORE INFORMATION
✓ **Answer**: C
## CONCEPT FRAMEWORK
	**Primary Concept**: Conversion of Galvanometer to Voltmeter and Ammeter
	**Related Topics**: Galvanometer, Voltmeter, Ammeter, Ohm's Law, Kirchhoff's Laws
	**Prerequisites**: Understanding of circuit elements and basic circuit analysis.
## SOLUTION PATHWAY
**Given Information**:
	$R_g = 50 \Omega$ (Galvanometer Resistance)
	$I_g = 100 \mu A = 100 	imes 10^{-6} A$ (Full-scale deflection current)
	$V = 5V$ (Voltmeter Range)
	$I = 10mA = 10 	imes 10^{-3} A$ (Ammeter Range)
**Method & Steps**:
	1. **Voltmeter Conversion:**
		Reasoning: To convert a galvanometer into a voltmeter, a high resistance ($R_x$) is connected in series such that the desired full-scale voltage corresponds to the galvanometer's full-scale current.
		Formula/Rule: $V = I_g(R_g + R_x)$
	2. **Ammeter Conversion:**
		Working: To convert a galvanometer into an ammeter, a low resistance ($R_s$) is connected in parallel (shunt) such that the excess current bypasses the galvanometer.
		Key Point: The voltage across the shunt resistor and the galvanometer is the same.
		Formula/Rule: $I_g R_g = (I - I_g)R_s$
	3. **Relationship between $R_x$ and $R_s$**
		Working: From the voltmeter equation, $R_x = \frac{V}{I_g} - R_g = \frac{5}{100 	imes 10^{-6}} - 50 = 50000 - 50 = 49950 \Omega$. From the ammeter equation, $R_s = \frac{I_g R_g}{I - I_g} = \frac{100 	imes 10^{-6} 	imes 50}{10 	imes 10^{-3} - 100 	imes 10^{-6}} = \frac{5 	imes 10^{-3}}{9.9 	imes 10^{-3}} \approx 0.505 \Omega$.
		Result: $R_x R_s = 49950 	imes 0.505 \approx 25224.75 \approx 25000$. Also, $10R_g^2 = 10 	imes (50)^2 = 25000$.
		Verification: Therefore, $R_x R_s \approx 10 R_g^2$.
 ## OPTION ANALYSIS
[A] R_x = R_s :x: Clearly, $R_x$ and $R_s$ are not equal.
[B] R_x = 1000 R_s :x: $R_x$ is much larger than $1000R_s$.
[C] R_x R_s ≈ 10 R_g^2 ✓: This is the correct relationship as derived above.
[D] R_x R_s = R_g^2 :x: This relationship is incorrect; the factor of 10 is missing.
 ## LEARNING AIDS
	**Common Mistakes**:
		1. Incorrectly applying Ohm's law: Ensure correct voltage and current values are used in calculations.
		2. Confusing series and parallel connections: Remember that for a voltmeter, the resistance is added in series, and for an ammeter, the resistance is added in parallel.
	**Quick Tips**:
		1. For voltmeter conversion, $R_x$ is much larger than $R_g$.
		2. For ammeter conversion, $R_s$ is much smaller than $R_g$.

VENKATESWARARAO · Answer

R_x = R_s

VENKATESWARARAO · Answer

R_x = 1000 R_s

VENKATESWARARAO · Answer

The correct Option is "R_x R_s ≈ 10 R_g^2"

VENKATESWARARAO · Answer

R_x R_s = R_g^2

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