Log In Start studying!

Select your language

Suggested languages for you:
Answers without the blur. Sign up and see all textbooks for free! Illustration


College Physics (Urone)
Found in: Page 777

Answers without the blur.

Just sign up for free and you're in.


Short Answer

A certain ammeter has a resistance of \(5.00 \times {10^{ - 5}}{\rm{ }}\Omega \) on its \(3.00 - A\) scale and contains a \(10.0 - \Omega \) galvanometer. What is the sensitivity of the galvanometer?

The galvanometer inside an ammeter with internal resistance \(r = 10{\rm{ }}\Omega \) has a sensitivity value of \({I_g} = 15{\rm{ }}\mu A\).

See the step by step solution

Step by Step Solution

Step 1: Concept Introduction

A galvanometer is an electromechanical device used to detect electric current. A galvanometer deflects a pointer in response to an electric current flowing through a coil in a constant magnetic field. An example of an actuator is a galvanometer.

The total flow of electrons via a wire can be used to describe the rate of electron flow. Anything that prevents current flow is referred to as "resistance." An electrical circuit needs resistance in order to transform electrical energy into light, heat, or movement.

Step 2: Information Provided

  • An ammeter with resistance: \(R = 5 \cdot {10^{ - 5}}{\rm{ }}\Omega \)
  • An ammeter with a scale: \({I_{tot}} = 3{\rm{ }}A\)
  • A galvanometer with internal resistance: \(r = 10{\rm{ }}\Omega \)

Step 3: Calculation for the sensitivity of galvanometer

The sensitivity of the galvanometer is the current that goes through the galvanometer\({I_g}\). Obtain the sensitivity by noting that \(R\) and \(r\) are at same potentials, which gives –

\(\begin{align}{c}{I_g}r & = {I_R}R\\{I_g}r & = \left( {{I_{tot}} - {I_g}} \right)R\\{I_g}(r + R) & = {I_{tot}}R\\{I_g} & = \frac{R}{{r + R}}{I_{tot}}\end{align}\)

\(\begin{align}{} & = \frac{{5\cot {{10}^{ - 5}}{\rm{ }}\Omega }}{{\left( {10 + 5 \cdot {{10}^{ - 5}}} \right)\Omega }}(3\;A)\\ & = 1.5 \cdot {10^{ - 5}}\;A\\ &= 15{\rm{ }}\mu A\end{align}\)

Therefore, the value of the current is obtained as \({I_g} = 15{\rm{ }}\mu A\).

Most popular questions for Physics Textbooks


Want to see more solutions like these?

Sign up for free to discover our expert answers
Get Started - It’s free

Recommended explanations on Physics Textbooks

94% of StudySmarter users get better grades.

Sign up for free
94% of StudySmarter users get better grades.