Question:
A moving coil galvanometer has a resistance of $50\,\Omega$ and gives a full-scale deflection for a current of $2 \text{ mA}$. To convert it into a voltmeter reading up to 10 V, the required series resistance to be connected is
A moving coil galvanometer has a resistance of $50\,\Omega$ and gives a full-scale deflection for a current of $2 \text{ mA}$. To convert it into a voltmeter reading up to 10 V, the required series resistance to be connected is
Show Hint
Think of it this way: Total resistance needed is $\frac{10\text{V}}{2\text{mA}} = 5000\,\Omega$. Since the galvanometer already provides $50\,\Omega$, you just need to add the remaining $5000 - 50 = 4950\,\Omega$ in series!
Updated On: Jun 3, 2026
- $4950\,\Omega$
- $5000\,\Omega$
- $4450\,\Omega$
- $5050\,\Omega$
Show Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Step 1: Concept
To convert a galvanometer into a voltmeter, a high resistance ($R$) must be connected in series with the galvanometer coil. The required value is given by the formula $R = \frac{V}{I_g} - G$, where $V$ is the maximum voltage to be measured, $I_g$ is the full-scale deflection current, and $G$ is the galvanometer resistance.
Step 2: Meaning
This formula comes from Ohm's law applied across the entire system: $V = I_g(G + R)$, where the total resistance is the sum of the series resistor and the internal coil resistance.
Step 3: Analysis
Given values: $G = 50\,\Omega$, $I_g = 2 \text{ mA} = 2 \times 10^{-3} \text{ A}$, and $V = 10 \text{ V}$. Substituting these values into the series resistance formula: $R = \frac{10}{2 \times 10^{-3}} - 50 = \frac{10000}{2} - 50 = 5000 - 50 = 4950\,\Omega$.
Step 4: Conclusion
Therefore, a series resistance of $4950\,\Omega$ must be connected to convert the galvanometer into the desired voltmeter.
Final Answer: (A)
To convert a galvanometer into a voltmeter, a high resistance ($R$) must be connected in series with the galvanometer coil. The required value is given by the formula $R = \frac{V}{I_g} - G$, where $V$ is the maximum voltage to be measured, $I_g$ is the full-scale deflection current, and $G$ is the galvanometer resistance.
Step 2: Meaning
This formula comes from Ohm's law applied across the entire system: $V = I_g(G + R)$, where the total resistance is the sum of the series resistor and the internal coil resistance.
Step 3: Analysis
Given values: $G = 50\,\Omega$, $I_g = 2 \text{ mA} = 2 \times 10^{-3} \text{ A}$, and $V = 10 \text{ V}$. Substituting these values into the series resistance formula: $R = \frac{10}{2 \times 10^{-3}} - 50 = \frac{10000}{2} - 50 = 5000 - 50 = 4950\,\Omega$.
Step 4: Conclusion
Therefore, a series resistance of $4950\,\Omega$ must be connected to convert the galvanometer into the desired voltmeter.
Final Answer: (A)
Was this answer helpful?
0
0
Top AP EAPCET Physics Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The radii of the two columns in a U - tube are r1 and r2. When a liquid of density p (angle of contact is $ 0{}^\circ $ ) is filled in it, the level difference of the liquid in the two arms is h. The surface tension of the liquid is: (g = acceleration due to gravity)
- A rigid metallic sphere is spinning around its own axis in the absence of external torque. If the temperature is raised, its volume increases by $9 \%$. The change in its angular speed is
View More Questions
Top AP EAPCET Basic Galvanometer Questions
- In a galvanometer, when the number of turns $N$ becomes $2N$ and resistance $R$ becomes doubled, then
- A galvanometer has a deflection of 25 divisions for a current of 0.1 A flowing through it. Then the current sensitivity of the galvanometer is
- A maximum current of 0.5 mA can pass through a galvanometer of resistance 15 \(\Omega\). The resistance to be connected in series to the galvanometer to convert it into a voltmeter of range 0–10 V is
- A moving coil galvanometer has 150 equal divisions. Its current sensitivity is 10 divisions per mA and Voltage sensitivity is 2 divisions per millivolt. In order to read each division 1 volt, the resistance in ohms to be connected in series is:
- A galvanometer has a coil of resistance \(100\,\Omega\) showing a full scale deflection at \(50\,\mu\text{A}\). The resistance that should be added to use it as an ammeter of range \(10\,\text{mA}\) is
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions