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Concepts Of Physics MCQ Edition [Volume 2]
Exam
Concepts Of Physics MCQ Edition [Volume 2]
Subject
Physics
Chapter
Electric Current in Conductors

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Electric Current in Conductors

A voltmeter coil possesses a resistance of 50.0 and is connected in...

A voltmeter coil possesses a resistance of 50.0Ω50.0 \Omega and is connected in series with a 1.15 kΩ1.15 \text{ k}\Omega resistor. It is capable of reading potential differences up to 12 V12\text{ V}. If the identical coil is utilized to build an ammeter that can measure currents up to 2.0 A2.0\text{ A}, determine the required resistance of the shunt.

Options

  1. A6.82Ω6.82 \Omega
  2. B0.251Ω0.251 \Omega
  3. C0.502Ω0.502 \Omega
  4. D6.03Ω6.03 \Omega

Correct answer

B. 0.251Ω0.251 \Omega

Step-by-step solution

Let RgR_g be the resistance of the coil and RsR_s be the series resistance.

Rg=50.0ΩR_g = 50.0 \Omega

Rs=1.15 kΩ=1150ΩR_s = 1.15 \text{ k}\Omega = 1150 \Omega

The maximum potential difference is V=12 VV = 12 \text{ V}.

The full-scale deflection current IgI_g is given by:

Ig=VRg+Rs=1250+1150=121200=0.01 AI_g = \dfrac{V}{R_g + R_s} = \dfrac{12}{50 + 1150} = \dfrac{12}{1200} = 0.01 \text{ A}

To convert this coil into an ammeter of range I=2.0 AI = 2.0 \text{ A}, a shunt resistance SS is connected in parallel.

S=IgRgI−IgS = \dfrac{I_g R_g}{I - I_g}

Substituting the values:

S=0.01×50.02.0−0.01=0.51.99≈0.251ΩS = \dfrac{0.01 \times 50.0}{2.0 - 0.01} = \dfrac{0.5}{1.99} \approx 0.251 \Omega
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