Answer
385.8k+ views
Hint If each division of the galvanometer were reading 1 volts, then the maximum volt readable by the galvanometer (full scale deflection) would be 150 V. Use the voltage and current sensitivity of the galvanometer to find the resistance of the galvanometer itself.
Formula used: In this solution we will be using the following formulae;
\[R = \dfrac{V}{I}\] where \[R\] is the resistance of a component, \[V\] is the potential difference across the component, and \[I\] is the current flowing through the component.
\[{R_{eqs}} = {R_1} + {R_2}\] where \[{R_{eqs}}\] is the equivalent resistance of two resistances in series, and \[{R_1},{R_2}\] are the individual resistances.
Complete Step-by-Step solution:
Now, the moving galvanometer has 150 equal divisions with 10 divisions per milli-ampere. Hence, the maximum current that can be measured by the galvanometer this way is
\[{I_g} = \dfrac{{150div}}{{10div/mA}} = 15mA\]
And similarly, the maximum voltage which can be measured by the galvanometer is
\[{V_g} = \dfrac{{150div}}{{2div/mA}} = 75mV\].
With this maximum, we can calculate the resistance of the galvanometer itself using the formula
\[R = \dfrac{V}{I}\] where \[R\] is the resistance of a component, \[V\] is the potential difference across the component, and \[I\] is the current flowing through the component.
Hence,
\[R = \dfrac{{{V_g}}}{{{I_g}}} = \dfrac{{75mV}}{{15mA}} = 5\Omega \]
Now, we want the galvanometer to read 1 volt per division, which means that the maximum should be
\[{V_{g2}} = 150div \times 1\dfrac{V}{{div}} = 150V\]
Hence, the resistance of the galvanometer should be
\[{R_{g2}} = \dfrac{{150A}}{{0.015A}} = 10000\Omega \]
Hence, since the resistance added in series should be
\[{R_s} = 10000 - 5 = 9995\Omega \]
Hence, the correct option is D
Note: For clarity, we subtract to get the resistance above because, 1000 ohms is the resistance the galvanometer should have but the galvanometer has 5 ohms. Hence, since the series equivalent resistance is given by
\[{R_{eqs}} = {R_1} + {R_2}\] where \[{R_{eqs}}\] is the equivalent resistance of two resistances in series, and \[{R_1},{R_2}\] are the individual resistances.
Hence,
\[10000\Omega = {R_s} + 5\Omega \]
\[ \Rightarrow {R_s} = 10000 - 5 = 9995\Omega \]
Formula used: In this solution we will be using the following formulae;
\[R = \dfrac{V}{I}\] where \[R\] is the resistance of a component, \[V\] is the potential difference across the component, and \[I\] is the current flowing through the component.
\[{R_{eqs}} = {R_1} + {R_2}\] where \[{R_{eqs}}\] is the equivalent resistance of two resistances in series, and \[{R_1},{R_2}\] are the individual resistances.
Complete Step-by-Step solution:
Now, the moving galvanometer has 150 equal divisions with 10 divisions per milli-ampere. Hence, the maximum current that can be measured by the galvanometer this way is
\[{I_g} = \dfrac{{150div}}{{10div/mA}} = 15mA\]
And similarly, the maximum voltage which can be measured by the galvanometer is
\[{V_g} = \dfrac{{150div}}{{2div/mA}} = 75mV\].
With this maximum, we can calculate the resistance of the galvanometer itself using the formula
\[R = \dfrac{V}{I}\] where \[R\] is the resistance of a component, \[V\] is the potential difference across the component, and \[I\] is the current flowing through the component.
Hence,
\[R = \dfrac{{{V_g}}}{{{I_g}}} = \dfrac{{75mV}}{{15mA}} = 5\Omega \]
Now, we want the galvanometer to read 1 volt per division, which means that the maximum should be
\[{V_{g2}} = 150div \times 1\dfrac{V}{{div}} = 150V\]
Hence, the resistance of the galvanometer should be
\[{R_{g2}} = \dfrac{{150A}}{{0.015A}} = 10000\Omega \]
Hence, since the resistance added in series should be
\[{R_s} = 10000 - 5 = 9995\Omega \]
Hence, the correct option is D
Note: For clarity, we subtract to get the resistance above because, 1000 ohms is the resistance the galvanometer should have but the galvanometer has 5 ohms. Hence, since the series equivalent resistance is given by
\[{R_{eqs}} = {R_1} + {R_2}\] where \[{R_{eqs}}\] is the equivalent resistance of two resistances in series, and \[{R_1},{R_2}\] are the individual resistances.
Hence,
\[10000\Omega = {R_s} + 5\Omega \]
\[ \Rightarrow {R_s} = 10000 - 5 = 9995\Omega \]
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)