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JEE Main 202315 AprilMorning Shift
Exam
JEE Main
Year
2023
Date
15 April
Shift
Morning Shift
Subject
Physics
Chapter
Current Electricity

JEE Main 2023 · 15 April · Morning Shift · Physics · Current Electricity

For designing a voltmeter of range 50V and an ammeter of range 10mA...

For designing a voltmeter of range 50V50V and an ammeter of range 10mA10mA using a galvanometer which has a coil of resistance 54Ω54 Ω showing a full scale deflection for 1mA1mA as in figure.

For designing a voltmeter of range 50V and an ammeter of range 10mA using a galvanometer which has a coil of resistance

(A) for voltmeter R≈50kΩR \approx 50k Ω

(B) for ammeter r≈0.2Ωr \approx 0 . 2 Ω

(C) for ammeter r≈6Ωr \approx 6 Ω

(D) for voltmeter R≈5kΩR \approx 5k Ω

(E) for voltmeter R≈500ΩR \approx 500 Ω

Choose the correct answer from the options given below:

Options

  1. A(A)\left(A\right) and (C)\left(C\right)
  2. B(C)\left(C\right) and (E)\left(E\right)
  3. C(C)\left(C\right) and (D)\left(D\right)
  4. D(A)\left(A\right) and (B)\left(B\right)

Correct answer

A. (A)\left(A\right) and (C)\left(C\right)

Step-by-step solution

The resistance is given by

R=VIg−GR = \dfrac{V}{{I}_{g}} - G

⇒5010−3−54≈50kΩ⇒ \dfrac{50}{{10}^{- 3}} - 54 \approx 50kΩ

For ammeter,

S=IgGI−Ig⇒S=10−3×54(10−1)×10−3=6ΩS = \dfrac{{I}_{g}G}{I - {I}_{g}} ⇒ S = \dfrac{{10}^{- 3} \times 54}{( 10 - 1 ) \times {10}^{- 3}} = 6Ω

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