
A galvanometer has a coil of resistance 100 ohm and gives a full scale deflection for 30 mA current. If it is to work as a voltmeter of 30 volt range, the resistance required to be added will be
\[
A.{\text{ }}1000\Omega \\
B.{\text{ }}900\Omega \\
C.{\text{ }}1800\Omega \\
D.{\text{ }}500\Omega \\
\]
Answer
609k+ views
- Hint: In order to deal with this question we will use the concept to convert the galvanometer into a voltmeter of a given range by connecting a resistance R in series of appropriate resistance according to the problem. Further we will use the formula to find the value of R which is mentioned in solution to get the required answer.
Formula used- $V = {I_g}\left( {G + R} \right)$
Complete step-by-step solution -
Given that
Resistance of galvanometer, $G = 100Q$
Current for full scale deflection:
${I_g} = 30mA = 30 \times {10^{ - 3}}A$
Range of voltmeter, $V = 30V$
To convert the galvanometer into a voltmeter of a given range, a resistance R is connected in series with it as shown in the figure.
From the above figure the voltage V across the points is:
$V = {I_g}\left( {G + R} \right)$
The above equation can be written as:
\[
\because V = {I_g}\left( {G + R} \right) \\
\Rightarrow G + R = \dfrac{V}{{{I_g}}} \\
\Rightarrow R = \dfrac{V}{{{I_g}}} - G \\
\]
Substitute the given values in above formula we have
\[
\because R = \dfrac{V}{{{I_g}}} - G \\
\Rightarrow R = \dfrac{{30}}{{30 \times {{10}^{ - 3}}}} - 100\Omega \\
\Rightarrow R = 1000\Omega - 100\Omega \\
\Rightarrow R = 900\Omega \\
\]
Hence, the resistance required to be added will be \[900\Omega \]
So, the correct answer is option B.
Note- A galvanometer is an electromechanical device which is used to measure and signify electrical current. A galvanometer acts as an actuator in response to electrical current flowing through a coil in a continuous magnetic field, by generating a rotary deflection (of a "pointer"). Galvanometer is an instrument which is used by deflection of a moving coil to measure a specific electrical current or a current operation. The deflection is a mechanical rotation which is derived from the current forces.
Formula used- $V = {I_g}\left( {G + R} \right)$
Complete step-by-step solution -
Given that
Resistance of galvanometer, $G = 100Q$
Current for full scale deflection:
${I_g} = 30mA = 30 \times {10^{ - 3}}A$
Range of voltmeter, $V = 30V$
To convert the galvanometer into a voltmeter of a given range, a resistance R is connected in series with it as shown in the figure.
From the above figure the voltage V across the points is:
$V = {I_g}\left( {G + R} \right)$
The above equation can be written as:
\[
\because V = {I_g}\left( {G + R} \right) \\
\Rightarrow G + R = \dfrac{V}{{{I_g}}} \\
\Rightarrow R = \dfrac{V}{{{I_g}}} - G \\
\]
Substitute the given values in above formula we have
\[
\because R = \dfrac{V}{{{I_g}}} - G \\
\Rightarrow R = \dfrac{{30}}{{30 \times {{10}^{ - 3}}}} - 100\Omega \\
\Rightarrow R = 1000\Omega - 100\Omega \\
\Rightarrow R = 900\Omega \\
\]
Hence, the resistance required to be added will be \[900\Omega \]
So, the correct answer is option B.
Note- A galvanometer is an electromechanical device which is used to measure and signify electrical current. A galvanometer acts as an actuator in response to electrical current flowing through a coil in a continuous magnetic field, by generating a rotary deflection (of a "pointer"). Galvanometer is an instrument which is used by deflection of a moving coil to measure a specific electrical current or a current operation. The deflection is a mechanical rotation which is derived from the current forces.
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