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A galvanometer having a resistance of 50...

A galvanometer having a resistance of `50Omega` , gives a full scale deflection for a current if `0.05` A. The length in metre if a resistance wire of area of cross- section `2.97 xx10^-2cm^2` that can be used to convert the gaivanometer into an ammeter which can read a maximum of 5 A current is (specific resistance of the wire =`5xx10^-7Omega-m`)

A

9

B

6

C

3

D

1.5

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem step-by-step, we need to determine the length of the resistance wire required to convert the galvanometer into an ammeter that can read a maximum current of 5 A. ### Step 1: Calculate the Shunt Resistance (S) The formula for the shunt resistance \( S \) needed to convert a galvanometer into an ammeter is given by: \[ S = \frac{I_g \cdot G}{I - I_g} \] Where: - \( I_g = 0.05 \, A \) (full-scale deflection current of the galvanometer) - \( G = 50 \, \Omega \) (resistance of the galvanometer) - \( I = 5 \, A \) (maximum current to be measured) Substituting the values: \[ S = \frac{0.05 \cdot 50}{5 - 0.05} \] Calculating the denominator: \[ 5 - 0.05 = 4.95 \] Now substituting back: \[ S = \frac{2.5}{4.95} \approx 0.505 \, \Omega \] ### Step 2: Use the Shunt Resistance to Find the Length of the Wire The resistance \( R \) of the wire can be calculated using the formula: \[ R = \frac{\rho \cdot l}{A} \] Where: - \( \rho = 5 \times 10^{-7} \, \Omega \cdot m \) (specific resistance of the wire) - \( A = 2.97 \times 10^{-2} \, cm^2 = 2.97 \times 10^{-6} \, m^2 \) (cross-sectional area converted to m²) - \( l \) is the length of the wire we need to find. Rearranging the formula to solve for \( l \): \[ l = \frac{R \cdot A}{\rho} \] Substituting the values: \[ l = \frac{0.505 \cdot 2.97 \times 10^{-6}}{5 \times 10^{-7}} \] Calculating the numerator: \[ 0.505 \cdot 2.97 \times 10^{-6} \approx 1.5 \times 10^{-6} \] Now substituting back into the equation: \[ l = \frac{1.5 \times 10^{-6}}{5 \times 10^{-7}} = 3 \, m \] ### Final Answer The length of the resistance wire required is approximately **3 meters**. ---
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Knowledge Check

  • A galvanometer having a resistance of 50 Omega , gives a full scale deflection for a current of 0.05 A. The length (in metres) of a resistance wire of area of cross section 3 xx 10^(-2) cm? that can be used to convert the galvanometer into an ammeter which can read a maximum of 5 A current is ("Specific resistance of the wirep "= 5 xx 10^(-7)Omega m)

    A
    9
    B
    6
    C
    3
    D
    1.5
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