A galvanometer has a coil of resistance \(200 \, \Omega\) with a full scale deflection at \(20 \, \mu A\). The value of resistance to be added to use it as an ammeter of range (0–20) mA is:
- \(0.40 \, \Omega\)
- \(0.20 \, \Omega\)
- \(0.50 \, \Omega\)
- \(0.10 \, \Omega\)
The Correct Option is B
Approach Solution - 1
To convert a galvanometer into an ammeter, a low resistance called a "shunt" is connected in parallel with the galvanometer. The shunt resistance allows the ammeter to measure larger currents by diverting most of the current around the galvanometer.
Given:
- Resistance of the galvanometer coil, \(R_g = 200 \, \Omega\)
- Full scale deflection current of the galvanometer, \(I_g = 20 \, \mu A = 20 \times 10^{-6} \, A\)
- Desired range of the ammeter, \(I = 20 \, mA = 20 \times 10^{-3} \, A\)
The current flowing through the shunt resistor, \(I_s\), is given by:
\(I_s = I - I_g\)
Substitute the values:
\(I_s = 20 \times 10^{-3} - 20 \times 10^{-6} = 19.98 \times 10^{-3} \, A\)
The voltage across the galvanometer and the shunt is the same. Therefore, we have:
\(I_g \times R_g = I_s \times R_s\)
We can rearrange to find the shunt resistance, \(R_s\):
\(R_s = \frac{I_g \times R_g}{I_s}\)
Substitute the values:
\(R_s = \frac{20 \times 10^{-6} \times 200}{19.98 \times 10^{-3}}\)
Calculate the value:
\(R_s = \frac{4 \times 10^{-3}}{19.98 \times 10^{-3}} \approx 0.20 \, \Omega\)
Therefore, the value of the resistance to be added so the galvanometer can be used as an ammeter with a range of 0-20 mA is \(0.20 \, \Omega\).
The correct answer is: \(0.20 \, \Omega\).
Approach Solution -2
Step 1: Formula for shunt resistance The shunt resistance \( R_s \) is given by:
\[ R_s = \frac{I_g R_g}{I - I_g}, \]
where:
- \( I_g = 20 \, \mu A = 20 \times 10^{-6} \, \text{A} \) (full-scale deflection current),
- \( R_g = 200 \, \Omega \) (resistance of the galvanometer),
- \( I = 20 \, mA = 20 \times 10^{-3} \, \text{A} \) (ammeter range).
Step 2: Substitute the values
\[ R_s = \frac{(20 \times 10^{-6}) \cdot 200}{(20 \times 10^{-3}) - (20 \times 10^{-6})}. \]
Simplify the numerator:
\[ (20 \times 10^{-6}) \cdot 200 = 4 \times 10^{-3}. \]
Simplify the denominator:
\[ (20 \times 10^{-3}) - (20 \times 10^{-6}) = 19.98 \times 10^{-3}. \]
Thus:
\[ R_s = \frac{4 \times 10^{-3}}{19.98 \times 10^{-3}} \approx 0.20 \, \Omega. \]
Final Answer: \( 0.20 \, \Omega. \)
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