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: \( 0.20 \, \Omega. \)
Top Questions on The Moving Coil Galvanometer
Assertion (A): The deflection in a galvanometer is directly proportional to the current passing through it.
Reason (R): The coil of a galvanometer is suspended in a uniform radial magnetic field.
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Galvanometer:
A galvanometer is an instrument used to show the direction and strength of the current passing through it. In a galvanometer, a coil placed in a magnetic field experiences a torque and hence gets deflected when a current passes through it.
The name "galvanometer" is derived from the surname of Italian scientist Luigi Galvani, who in 1791 discovered that electric current makes a dead frog’s leg jerk.
A spring attached to the coil provides a counter torque. In equilibrium, the deflecting torque is balanced by the restoring torque of the spring, and we have the relation:
\[ NBAI = k\phi \]
Where:
- \( N \) is the total number of turns in the coil
- \( A \) is the area of cross-section of each turn
- \( B \) is the radial magnetic field
- \( k \) is the torsional constant of the spring
- \( \phi \) is the angular deflection of the coil
As the current \( I_g \) that produces full-scale deflection in the galvanometer is very small, the galvanometer alone cannot be used to measure current in electric circuits.
To convert a galvanometer into an ammeter (to measure larger currents), a small resistance called a shunt is connected in parallel to the galvanometer.
To convert it into a voltmeter (to measure potential difference), a high resistance is connected in series with the galvanometer.
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