Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Current Electricity question

2019 · 12 Jan · Shift 2 · Q65
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Physics
  4. /Current Electricity
  5. /2019 · 12 Jan · Shift 2 · Q65

Current Electricity question

2019 · 12 Jan · Shift 2 · Q65

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
A galvanometer, whose resistance is 50 ohm, has 25 divisions in it. When a current of 4 ×\times× 10–4 A passes through it, its needle ( pointer) deflects by one division. To use this galvanometer as a voltmeter of range 2.5 V, it should be connected to a resistance of :
  1. A
    200 ohm
  2. B
    250 ohm
  3. C
    6200 ohm
  4. D
    6250 ohm
View written solutionFree

Correct answer: A

  1. Given data
  • Galvanometer resistance: G=50 ΩG = 50\,\OmegaG=50Ω
  • Total divisions on scale: 252525
  • Current for 1 division deflection: 4×10−4 A4 \times 10^{-4}\,\text{A}4×10−4A
  • Required voltmeter range: V=2.5 VV = 2.5\,\text{V}V=2.5V
  1. Find the full-scale deflection current

Since 1 division requires current 4×10−4 A4 \times 10^{-4}\,\text{A}4×10−4A, for 252525 divisions:

Ig=25×4×10−4I_g = 25 \times 4 \times 10^{-4}Ig​=25×4×10−4 Ig=100×10−4=10−2 A=0.01 AI_g = 100 \times 10^{-4} = 10^{-2}\,\text{A} = 0.01\,\text{A}Ig​=100×10−4=10−2A=0.01A

So, the galvanometer gives full-scale deflection at

Ig=0.01 AI_g = 0.01\,\text{A}Ig​=0.01A
  1. Convert galvanometer into a voltmeter

To make a voltmeter of range 2.5 V2.5\,\text{V}2.5V, a high resistance RRR is connected in series with the galvanometer.

At full-scale deflection,

V=Ig(G+R)V = I_g (G + R)V=Ig​(G+R)

Substitute the values:

2.5=0.01(50+R)2.5 = 0.01(50 + R)2.5=0.01(50+R)
  1. Solve for RRR
50+R=2.50.01=25050 + R = \frac{2.5}{0.01} = 25050+R=0.012.5​=250 R=250−50=200 ΩR = 250 - 50 = 200\,\OmegaR=250−50=200Ω
  1. Check the options
  • A: 200 Ω200\,\Omega200Ω ✅
  • B: 250 Ω250\,\Omega250Ω ❌
  • C: 6200 Ω6200\,\Omega6200Ω ❌
  • D: 6250 Ω6250\,\Omega6250Ω ❌

Hence, the required series resistance is

200 Ω\boxed{200\,\Omega}200Ω​
PreviousNext

More from Current Electricity

  • In a meter bridge, as shown in the figure, it is given that resistance Y=12.5Ω and that the balance is obtained at a distance 39.5cm from end A(by Jockey J). After interchanging the resistances X and Y, a new balance… Includes diagram2018 · MCQ
  • In the given circuit all resistances are of value Rohm each. The equivalent resistance between A and B is : Includes diagram2018 · MCQ
  • A copper rod of cross-sectional area A carries a uniform current I through it. At temperature T, if the volume charge density of the rod is ρ, how long will the changes take to travel a distance d ?2018 · MCQ
  • A constant voltages is applied between two ends of a metallic wire. If the length is halved and the radius of the wire is doubled, the rate of heat developed in the wire will be :2018 · MCQ
  • A galvanometer with its coil resistance 25 Ω requires a current of 1 mA for its full deflection. In order to construct an ammeter to read upto a current of 2 A, the approximate value of the shunt resistance should be :2018 · MCQ
  • A heating element has a resistance of 100 Ω at room temperature. When it is connected to a supply of 220 V, a steady current of 2 A passes in it and temperature is 500oC more than room temperature. what is the temperature…2018 · MCQ
  • In a circuit for finding the resistance of a galvanometer by half deflection method, a 6 V battery and a high resistance of 11 k Ω are used. The figure of merit of the galvanometer is 60 μA/ division. In the absence of shunt…2018 · MCQ
  • On interchanging the resistances, the balance point of a meter bridge shifts to the left by 10 cm. The resistance of their series combination is 1 k Ω. How much was the resistance on the left slot before interchanging the…2018 · MCQ