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A moving-coil instrument gives a full-scale deflection if 20 mA flows through it. The resistance of the coil is 2 . Calculate:

Physics

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2 parts
a

0.0408 Ω\text{}0.0408 \text{ } \Omega\text{}

b

498 Ω\text{}498 \text{ } \Omega\text{}

Step-by-step solution

4 steps

Here are the solutions for Assessment activity 12.6.

1. A moving-coil instrument gives a full-scale deflection if 20 mA flows through it. The resistance of the coil is 2 Ω\Omega. Calculate:

Given:

  • Meter full-scale deflection current, Im=20 mA=0.020 AI_m = 20 \text{ mA} = 0.020 \text{ A}
  • Meter coil resistance, Rm=2 ΩR_m = 2 \text{ } \Omega

Part a· 2 steps

the resistance required to use the instrument as an ammeter for a 1 A current flow

  1. Determine the current through the shunt resistor.

    To use the instrument as an ammeter, a shunt resistor (RshR_{sh}) is connected in parallel with the meter. The total current (II) is 1 A1 \text{ A}.
    The current flowing through the shunt resistor is the total current minus the current flowing through the meter:

    Ish=I−ImI_{sh} = I - I_m Ish=1 A−0.020 AI_{sh} = 1 \text{ A} - 0.020 \text{ A} Ish=0.980 AI_{sh} = 0.980 \text{ A}
  2. Calculate the required shunt resistance.

    Since the shunt resistor is in parallel with the meter, the voltage across them is the same (Vsh=VmV_{sh} = V_m).

    IshRsh=ImRmI_{sh} R_{sh} = I_m R_m Rsh=ImRmIshR_{sh} = \frac{I_m R_m}{I_{sh}}

    Substitute the values:

    Rsh=(0.020 A)(2 Ω)0.980 AR_{sh} = \frac{(0.020 \text{ A})(2 \text{ } \Omega)}{0.980 \text{ A}} Rsh=0.040 Ω0.980R_{sh} = \frac{0.040 \text{ } \Omega}{0.980} Rsh≈0.0408 ΩR_{sh} \approx 0.0408 \text{ } \Omega

    The required shunt resistance is 0.0408 Ω\boxed{\text{}0.0408 \text{ } \Omega\text{}}

Part b

498 Ω\text{}498 \text{ } \Omega\text{}

the series resistance required to use the instrument as a voltmeter up to 10 V.

  1. Determine the total resistance needed for the voltmeter.

    To use the instrument as a voltmeter, a series resistor (RseR_{se}) is connected in series with the meter. The maximum voltage (VV) to be measured is 10 V10 \text{ V}.
    At full-scale deflection, the total resistance (RtotalR_{total}) of the voltmeter circuit is given by Ohm's law:

    Rtotal=VImR_{total} = \frac{V}{I_m} Rtotal=10 V0.020 AR_{total} = \frac{10 \text{ V}}{0.020 \text{ A}} Rtotal=500 ΩR_{total} = 500 \text{ } \Omega
  2. Calculate the required series resistance.

    The total resistance is the sum of the meter's resistance and the series resistance:

    Rtotal=Rm+RseR_{total} = R_m + R_{se} Rse=Rtotal−RmR_{se} = R_{total} - R_m

    Substitute the values:

    Rse=500 Ω−2 ΩR_{se} = 500 \text{ } \Omega - 2 \text{ } \Omega Rse=498 ΩR_{se} = 498 \text{ } \Omega

    The required series resistance is 498 Ω\boxed{\text{}498 \text{ } \Omega\text{}}

    2. What material are shunt resistors made of? Why?

    Shunt resistors are typically made of materials like Manganin or Constantan.

    These materials are chosen for shunt resistors primarily because they have a very low temperature coefficient of resistance. This means their resistance changes very little with variations in temperature, ensuring that the ammeter's accuracy is maintained even when the instrument heats up due to current flow. They also have high resistivity and low thermal electromotive force (EMF) against copper.

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