A galvanometer of resistance $50\Omega $ is connected to a battery of 3V along with a resistance of 2950$\Omega $ in series. A full-scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be:
A. 6050$\Omega $
B. 4450$\Omega $
C. 5050$\Omega $
D. 5550$\Omega $
Answer
662.1k+ views
- Hint: In such questions we need the concept of galvanometer and the resistance in series and parallel to be understood thoroughly. It is an electrochemical instrument and it is used for detecting and indicating an electric current. In this problem we need to calculate the equivalent resistance when the deflection is 20 divisions with the help of the given information of full scale deflection.
Formula used – 1) Resistance in series,\[{R_s} = {R_1} + {R_2}....\]
2) \[V = IR\]
Complete step-by-step solution -
Given, ${R_G} = 50\Omega $\[V = IR\]
As the galvanometer of resistance 50 Ω is connected to a battery along with a resistance 2950 Ω in series
Therefore, ${R_{eq}} = {R_1} + {R_G} = 2950 + 50 = 3000\Omega $
Now, current through galvanometer $I = \dfrac{V}{R} = \dfrac{3}{{3000}} = {10^{ - 3}}A$
Current for 30 divisions = $I = {10^{ - 3}}A$
⇒current for 1 division = $\dfrac{I}{{30}} = \dfrac{{{{10}^{ - 3}}}}{{30}} = \dfrac{1}{3} \times {10^{ - 4}}A$
⇒current for 20 divisions =$\dfrac{{20I}}{{30}} = \dfrac{{20 \times {{10}^{ - 3}}}}{{30}} = \dfrac{2}{3} \times {10^{ - 3}}A$
Now, equivalent resistance will be , ${R_{eq}} = \dfrac{{3V \times 3}}{{(2 \times {{10}^{ - 3}})}} = \dfrac{{9000}}{2} = 4500\Omega $
As it is given, resistance of galvanometer is 50 Ω
So, resistance in series should be, \[R = {\text{ }}{R_{eq}}{\text{ }} - \;{R_G}\]= 4500 - 50 = 4450 Ω
Hence, the correct answer is 4450 Ω
Hence, the correct option is B.
Note – Galvanometer is a device used to check the flow of current in the circuit and it is in series combination with the series likewise ammeter. A galvanometer works as an actuator, by producing a rotary deflection (of a "pointer"), in response to electric current flowing through a coil in a constant magnetic field.
Formula used – 1) Resistance in series,\[{R_s} = {R_1} + {R_2}....\]
2) \[V = IR\]
Complete step-by-step solution -
Given, ${R_G} = 50\Omega $\[V = IR\]
As the galvanometer of resistance 50 Ω is connected to a battery along with a resistance 2950 Ω in series
Therefore, ${R_{eq}} = {R_1} + {R_G} = 2950 + 50 = 3000\Omega $
Now, current through galvanometer $I = \dfrac{V}{R} = \dfrac{3}{{3000}} = {10^{ - 3}}A$
Current for 30 divisions = $I = {10^{ - 3}}A$
⇒current for 1 division = $\dfrac{I}{{30}} = \dfrac{{{{10}^{ - 3}}}}{{30}} = \dfrac{1}{3} \times {10^{ - 4}}A$
⇒current for 20 divisions =$\dfrac{{20I}}{{30}} = \dfrac{{20 \times {{10}^{ - 3}}}}{{30}} = \dfrac{2}{3} \times {10^{ - 3}}A$
Now, equivalent resistance will be , ${R_{eq}} = \dfrac{{3V \times 3}}{{(2 \times {{10}^{ - 3}})}} = \dfrac{{9000}}{2} = 4500\Omega $
As it is given, resistance of galvanometer is 50 Ω
So, resistance in series should be, \[R = {\text{ }}{R_{eq}}{\text{ }} - \;{R_G}\]= 4500 - 50 = 4450 Ω
Hence, the correct answer is 4450 Ω
Hence, the correct option is B.
Note – Galvanometer is a device used to check the flow of current in the circuit and it is in series combination with the series likewise ammeter. A galvanometer works as an actuator, by producing a rotary deflection (of a "pointer"), in response to electric current flowing through a coil in a constant magnetic field.
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