The Moving Coil Galvanometer

Answers & Solutions

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

1. 1.↑ back to the question

   Current sensitivity is defined as the deflection of the coil per unit current flowing in it, i.e. $I_s = \dfrac{\theta}{I} = \dfrac{NAB}{k}$ where $N$ is the number of turns, $A$ the area, $B$ the magnetic field strength, and $k$ the torsional constant of the suspension.

2. 2.↑ back to the question

   When a current carrying coil is suspended in a uniform magnetic field, a torque acts on it, the magnitude of which depends on the strength of current. This torque tends to rotate the coil about the axis of suspension, so that the magnetic flux passing through the coil is maximum.

3. 3.↑ back to the question

   Let the shunt resistance be $R$. So, $V_{shunt} = V_{ammeter}$ (parallel combination), giving: $1\times0.8 = (5-1)\times R$ $0.8 = 4R \Rightarrow R = 0.2\ \Omega$

4. 4.↑ back to the question

   To convert it into a voltmeter, a large resistance $R$ is connected in series with the galvanometer. $\therefore R = \dfrac{V}{I_g} - R_G = \dfrac{3}{4\times10^{-3}} - 16$ $R = 750 - 16 = 734\ \Omega$

5. 5.↑ back to the question

   (a) The current sensitivity is given by $I_s = \dfrac{\theta}{I} = \dfrac{NAB}{k}$. Current sensitivity can be increased when: (i) number of turns $N$ is increased; (ii) area of coil $A$ is increased; (iii) magnetic field $B$ should be radial (and made as strong as possible, e.g. using a stronger permanent magnet). (b) $1\text{ V} = I_g(G+R_1)$ ...(i) $2\text{ V} = I_g(G+R_2)$ ...(ii) Dividing (ii) by (i): $\dfrac{2}{1} = \dfrac{G+R_2}{G+R_1}$ $\Rightarrow 2G + 2R_1 = G + R_2 \Rightarrow G = R_2 - 2R_1$

6. 6.↑ back to the question

   (a) Current sensitivity is defined as the deflection of coil per unit current flowing in it, i.e. $I_s = \dfrac{\theta}{I} = \dfrac{NAB}{k}$. (b)(i) A galvanometer can be converted into an ammeter of given range by connecting a suitable low resistance $S$ called shunt in parallel to the given galvanometer, whose value is given by $S = \left(\dfrac{I_g}{I_0-I_g}\right)G$ where $I_g$ is the current for full scale deflection of galvanometer, $I_0$ is the current to be measured by the galvanometer and $G$ is the resistance of galvanometer. (ii) The effective resistance of this ammeter (galvanometer $G$ parallel with shunt $S$) is $R_{eff} = \dfrac{GS}{G+S}$, which works out to $R_{eff} = \dfrac{G I_g}{I_0}$ (using $S = \dfrac{I_g G}{I_0-I_g}$, the parallel combination simplifies to this low value, always less than $S$ and $G$ individually).

7. 7.↑ back to the question

   (a) A galvanometer can be converted into an ammeter of given range by connecting a suitable low resistance $S$ called shunt in parallel to the given galvanometer, whose value is given by $S = \left(\dfrac{I_g}{I_0-I_g}\right)G$, where $I_g$ is the current for full scale deflection of galvanometer, $I_0$ is the current to be measured by the galvanometer and $G$ is the resistance of galvanometer. In order to increase the range of an ammeter $n$ times, the value of shunt resistance to be connected in parallel is $S = G/(n-1)$. (b) For $I_0 = 6$ A: $S = \left(\dfrac{I_g}{I_0-I_g}\right)G = \dfrac{4\times10^{-3}}{6-0.004}\times15$ $S \approx \dfrac{4\times10^{-3}}{6}\times15 = 0.01\ \Omega$

8. 8.↑ back to the question

   (a) As the ammeter is connected in series in the circuit, therefore, its resistance must be very low so that the circuit current is not affected. A very low shunt resistance makes the effective resistance of galvanometer (and hence the ammeter) very low. (b) $S = \dfrac{I_g G}{I-I_g} = \dfrac{6\times10^{-3}\times15}{6-6\times10^{-3}}$ $S = \dfrac{6\times10^{-3}\times15}{5.994} \approx 0.015\ \Omega$

9. 9.↑ back to the question

   (a) A galvanometer can be converted into a voltmeter of given range by connecting a suitable resistance $R$ in series with the galvanometer, whose value is given by $R = \dfrac{V}{I_g} - G$, where $V$ is the voltage to be measured, $I_g$ is the current for full scale deflection of galvanometer and $G$ is the resistance of galvanometer. Voltmeter is a high resistance instrument and it is always connected in parallel with the circuit element across which potential difference is to be measured. An ideal voltmeter has infinite resistance. In order to increase the range of voltmeter $n$ times, the value of resistance to be connected in series with galvanometer is $R = (n-1)G$. (b) $V_1 = I_g(R+980)$ and $V_2 = I_g(R+470)$ Also, $V_2 = V_1/2$ (given the range gets halved with the smaller series resistance) $\Rightarrow 2(R+470) = R+980 \Rightarrow 2R + 940 = R + 980 \Rightarrow R = 40\ \Omega$

10. 10.↑ back to the question

    Working principle: It works on the principle that a current-carrying coil placed in a magnetic field experiences a torque, the magnitude of which depends on the strength of current. (i) Radial magnetic field: To maintain the plane of the coil always parallel to the field $\vec{B}$ and to have maximum torque at every orientation of the coil (so that $\theta = 90°$ always and $\sin\theta = 1$ throughout the rotation), giving a uniform, linear response. (ii) A cylindrical soft iron core: This has high permeability and it intensifies the magnetic field and hence increases the sensitivity of the galvanometer. Current sensitivity, $I_S = \dfrac{\theta}{I} = \dfrac{NBA}{k}$ No, a galvanometer as such cannot be used directly for measuring current in amperes, because it is an extremely sensitive instrument (full-scale deflection occurs even for a very small current) and it does not have a calibrated current scale of its own; it can, however, be used to detect current or, with a shunt, converted into an ammeter of a specific range. By measuring the deflection produced in the galvanometer coil (relating $\theta$ to $I$ via its known sensitivity), one can in principle obtain the current, but not as a ready-made ammeter reading.

11. 11.↑ back to the question

    Principle: A current carrying coil placed in a magnetic field experiences a torque, the magnitude of which depends on the strength of current. Galvanometer as a voltmeter — Case (I): $R_1 = \dfrac{V}{I_g} - G$ ...(i) Case (II): $R_2 = \dfrac{V}{2I_g} - G$ ...(ii) Case (III), for range $2V$: $R_3 = \dfrac{2V}{I_g} - G$ ...(iii) $I_g$ = current through galvanometer which is fixed. From eqns. (i) and (ii): $R_1 - R_2 = \dfrac{V}{2I_g}$, so $G = R_1 - 2R_2$ Putting these values in eqn. (iii), we get $R_3 = 4(R_1-R_2) - (R_1-2R_2) = 3R_1 - 2R_2$

12. 12.↑ back to the question

    (a) Magnetic field is radial in moving coil galvanometer so that the plane of the coil always lies in the direction of the magnetic field (i.e. $\theta = 90°$ is maintained at every orientation), giving maximum and constant torque, so that the scale is linear. A radial magnetic field is produced by (i) properly cutting the magnetic pole pieces in the shape of concave faces, and (ii) using a soft iron core within the coil. (b) $V = (R+G)I_g$ ...(i) $V' = (R'+G)I_g$ ...(ii) From equations (i) and (ii): $\dfrac{V'}{V} = \dfrac{R'+G}{R+G} = \dfrac{1}{2} \Rightarrow R' = \dfrac{R}{2} - \dfrac{G}{2}$ (This gives the resistance needed to change the range from $0$ to $V$ down to $0$ to $V/2$, in terms of the original series resistance $R$ and galvanometer resistance $G$.)

13. 13.↑ back to the question

    (a) Working: when a current flows through the coil, a torque acts on it, $\tau = NIAB$ (taking $\sin\theta=1$ since the field is radial). This magnetic torque tends to rotate the coil, and a spring provides a counter (restoring) torque $k\phi$ that balances the magnetic torque, resulting in a steady angular deflection $\phi$. In equilibrium, $k\phi = NIAB \Rightarrow \phi = \left(\dfrac{NAB}{k}\right)I$, i.e. $\phi \propto I$. Since deflection is directly proportional to current, we get a linear scale. The uniform radial magnetic field keeps the plane of the coil always parallel to the direction of the magnetic field for every orientation, ensuring $\sin\theta = 1$ throughout the coil's rotation, which is what makes the scale linear (justifying its necessity). (b) Case (i): $V = I_g(G+2\text{k}\Omega)$ ...(i) Case (ii): $2V = I_g(G+5\text{k}\Omega)$ ...(ii) Dividing (ii) by (i): $2 = \dfrac{G+5000}{G+2000} \Rightarrow 2G+4000 = G+5000 \Rightarrow G = 1000\ \Omega$ From (i): $I_g = \dfrac{V}{G+2000} = \dfrac{V}{3000}$ For range $V/2$: let series resistance be $R$. $\dfrac{V}{2} = I_g(G+R) = \dfrac{V}{3000}(1000+R)$ $\Rightarrow 1500 = 1000+R \Rightarrow R = 500\ \Omega$

14. 14.↑ back to the question

    (i) Principle and working: A galvanometer is used to detect current in a circuit. When current $I$ is passed in the coil, torque $\tau$ acts on the coil, given by $\tau = NIAB\sin\theta$, where $\theta$ is the angle between the normal to plane of coil and the magnetic field of strength $B$, and $N$ is the number of turns in the coil. When the magnetic field is radial (as in the case of cylindrical pole pieces and soft iron core), then in every position of coil, the plane of the coil is parallel to the magnetic field lines, so that $\theta = 90°$ and $\sin90° = 1$. Deflecting torque, $\tau = NIAB$. If $C$ is the torsional rigidity of the wire and $\phi$ is the twist of suspension strip, then restoring torque $= C\phi$. For equilibrium, deflecting torque = restoring torque, i.e. $NIAB = C\phi \Rightarrow \phi = \left(\dfrac{NAB}{C}\right)I$, i.e. $\phi \propto I$. Deflection of coil is directly proportional to current flowing in the coil and hence we can construct a linear scale. (ii) The uniform radial magnetic field keeps the plane of the coil always parallel to the direction of the magnetic field, i.e. the angle between the plane of the coil and the magnetic field is zero (equivalently, $\theta=90°$ between normal and field) for all orientations of the coil. It is produced by using cylindrical pole pieces (concave in shape) together with a soft iron cylindrical core placed within the coil. (iii) Current sensitivity of a moving coil galvanometer is defined as the deflection of coil per unit current passed through it: $I_S = \dfrac{\theta}{I} = \dfrac{NBA}{k}$, where $N$ is the number of turns, $A$ is the area of the coil, $B$ is the magnetic field strength of the poles and $k$ is the spring (torsional) constant of the suspension wire. A convenient way to increase current sensitivity is to increase the number of turns $N$ (or the area $A$, or use a stronger magnet for larger $B$, or a suspension of lower torsional constant $k$).

15. 15.↑ back to the question

    (a) Principle and working: When current ($I$) is passed in the coil, torque $\tau$ acts on the coil, given by $\tau = NIAB\sin\theta$, where $\theta$ is the angle between the normal to plane of coil and the magnetic field of strength $B$, $N$ is the number of turns in a coil. When the magnetic field is radial, as in the case of cylindrical pole pieces and soft iron core, then in every position of coil, the plane of the coil is parallel to the magnetic field lines, so that $\theta = 90°$ and $\sin90° = 1$. Deflecting torque, $\tau = NIAB$. If $C$ is the torsional rigidity of the wire and $\theta$ is the twist of suspension strip, then restoring torque $= C\theta$. For equilibrium, deflecting torque = restoring torque, i.e. $NIAB = C\theta$. $\therefore \theta = \dfrac{NAB}{C}I$, i.e. $\theta \propto I$. Deflection of coil is directly proportional to current flowing in the coil, hence we can construct a linear scale. (i) The uniform radial magnetic field keeps the plane of the coil always parallel to the direction of the magnetic field, i.e. the angle between the plane of the coil and the magnetic field is zero for all orientations of the coil. (ii) The cylindrical soft iron core, when placed inside the coil of a galvanometer, makes the magnetic field stronger and radial in the space between it and the pole pieces, such that whatever be the position of the coil, the magnetic field is always parallel to its plane — this also increases sensitivity. (b) Current sensitivity: $I_S = \dfrac{NBA}{k}$. Voltage sensitivity, $V_S = \dfrac{NBA}{kR}$, where $R$ is the resistance of the wire (coil). Voltage sensitivity = Current sensitivity / $R$. Thus, on increasing the current sensitivity (e.g. by increasing $N$), voltage sensitivity may or may not increase, because increasing $N$ also increases the resistance $R$ of the coil by a similar (or greater) proportion, and the two changes can offset each other in the ratio $I_S/R$.

16. 16.Answer: (D)↑ back to the question

    Answer: (d) high resistance in series. As the ammeter is connected in place of a voltmeter, its resistance should be high (a voltmeter is ideally a high-resistance instrument). So, we have to connect a high resistance in series.

17. 17.Answer: (C)↑ back to the question

    Answer: (c) (A) is true, but (R) is false. When the resistance is lower, the range of ammeter is higher. So, Assertion (A) is true. To increase the range of ammeter, additional shunt is added in parallel (not in series) with the galvanometer. So, Reason (R) is false.