Examkin

Question

Difficulty: MediumElectric Circuits and Measuring Instruments

A galvanometer with an internal resistance of 40 Ω40\text{ }\Omega gives a full-scale deflection when a current of 10 mA10\text{ mA} passes through it. What resistance must be connected in series with the galvanometer to convert it into a voltmeter capable of measuring potential differences up to 10 V10\text{ V}?

  1. 960 Ω960\text{ }\OmegaAnswer
  2. B
    1000 Ω1000\text{ }\Omega
  3. C
    1040 Ω1040\text{ }\Omega
  4. D
    60 Ω60\text{ }\Omega

Answer

The required multiplier resistance is 960 Ω960\text{ }\Omega.
To convert a galvanometer into a voltmeter, a multiplier resistor RmR_m is connected in series. The total resistance of the voltmeter combination is Rtotal=Rg+Rm=VIg=10 V0.010 A=1000 ΩR_{\text{total}} = R_g + R_m = \frac{V}{I_g} = \frac{10\text{ V}}{0.010\text{ A}} = 1000\text{ }\Omega. Subtracting the galvanometer's internal resistance (40 Ω40\text{ }\Omega) yields Rm=960 ΩR_m = 960\text{ }\Omega.

Step-by-Step Solution

1
Convert the full-scale deflection current to amperes.
Ig=10 mA=10×10−3 A=0.010 AI_g = 10\text{ mA} = 10 \times 10^{-3}\text{ A} = 0.010\text{ A}.
Standard SI units must be used for electrical calculations.
2
Apply the voltmeter multiplier conversion formula.
V=Ig(Rg+Rm)  ⟹  Rm=VIg−RgV = I_g(R_g + R_m) \implies R_m = \frac{V}{I_g} - R_g.
The multiplier resistor RmR_m is connected in series with the galvanometer resistance RgR_g.
3
Substitute the known values into the equation.
Rm=100.010−40=1000−40=960 ΩR_m = \frac{10}{0.010} - 40 = 1000 - 40 = 960\text{ }\Omega.
Subtracting internal resistance gives the external resistance needed for full-scale voltage rating.

Key Concept

Voltmeter Conversion using a Series Multiplier Resistor
Rate this question

Topics

Electric Circuits and Measuring Instruments

More questions from this topic