The Moving Coil Galvanometer (Sensitivity) and Ammeter and Voltmeter Conversion Questions in English

(A) Given: Galvanometer resistance $G = 100\, \Omega$,full-scale current $I_g = 10\, mA = 10 \times 10^{-3}\, A$,and the desired range $I = 100\, mA = 100 \times 10^{-3}\, A$.
The formula for the shunt resistance $S$ required to convert a galvanometer into an ammeter is given by:
$S = \frac{I_g \cdot G}{I - I_g}$
Substituting the values:
$S = \frac{10 \times 10^{-3} \times 100}{100 \times 10^{-3} - 10 \times 10^{-3}}$
$S = \frac{1}{90 \times 10^{-3}} = \frac{1000}{90}$
$S = 11.11\, \Omega$.

(C) Given: Galvanometer resistance $G = 100 \, \Omega$,full-scale deflection current $I_g = 10^{-5} \, A$,and the desired range $I = 1 \, A$.
To convert a galvanometer into an ammeter,a shunt resistance $S$ is connected in parallel.
The formula for shunt resistance is $S = \frac{I_g \cdot G}{I - I_g}$.
Substituting the values: $S = \frac{10^{-5} \times 100}{1 - 10^{-5}}$.
Since $10^{-5}$ is negligible compared to $1$,we have $S \approx \frac{10^{-5} \times 10^2}{1} = 10^{-3} \, \Omega$.
Thus,the required shunt resistance is $10^{-3} \, \Omega$.

(D) Let $G$ be the resistance of the galvanometer and $S$ be the resistance of the shunt.
Given: $G = 36 \ \Omega$,$S = 4 \ \Omega$.
The fraction of the total current $i$ passing through the galvanometer is given by the formula:
$f_0 = \frac{i_g}{i} = \frac{S}{G + S}$
Substituting the given values:
$f_0 = \frac{4}{36 + 4} = \frac{4}{40} = \frac{1}{10}$
Therefore,the fraction of the total current passing through the galvanometer is $\frac{1}{10}$.

(B) The formula for the shunt resistance $S$ required to convert a galvanometer of resistance $G$ into an ammeter of range $I$ is given by $S = \frac{I_g G}{I - I_g}$.
Given:
Galvanometer resistance $G = 50\, \Omega$.
Full-scale deflection current $I_g = 0.01\, A$.
Desired range of ammeter $I = 10\, A$.
Substituting the values into the formula:
$S = \frac{0.01 \times 50}{10 - 0.01}$
$S = \frac{0.5}{9.99}$
$S \approx 0.05\, \Omega$.
Thus,the shunt resistance is $0.05\, \Omega$.

(A) Given:
Full-scale current of the ammeter,$I_g = 1\, A$.
Internal resistance of the ammeter,$G = 0.81\, \Omega$.
Desired range of the ammeter,$I = 10\, A$.
Let $S$ be the required shunt resistance.
The formula for the shunt resistance required to increase the range of an ammeter is given by:
$S = \frac{I_g \cdot G}{I - I_g}$
Substituting the given values:
$S = \frac{1 \cdot 0.81}{10 - 1}$
$S = \frac{0.81}{9}$
$S = 0.09\, \Omega$.
Therefore,the required shunt resistance is $0.09\, \Omega$.

(B) To convert a galvanometer into a voltmeter,a high resistance $R$ must be connected in series with it.
The formula for the series resistance is given by $R = \frac{V}{I_g} - G$,where $V$ is the maximum voltage to be measured,$I_g$ is the full-scale deflection current of the galvanometer,and $G$ is the resistance of the galvanometer.
Given: $V = 50 \,V$,$I_g = 10 \,mA = 10 \times 10^{-3} \,A = 0.01 \,A$,and $G = 40 \,\Omega$.
Substituting the values: $R = \frac{50}{0.01} - 40$.
$R = 5000 - 40 = 4960 \,\Omega$.
Thus,the required resistance is $4960 \,\Omega$.

(D) The full-scale deflection current $I_g$ is calculated by multiplying the number of divisions by the current per division: $I_g = 25 \times 4 \times 10^{-4} \, A = 100 \times 10^{-4} \, A = 0.01 \, A$.
To convert a galvanometer into a voltmeter, a high resistance $R$ must be connected in series with it.
The formula for the series resistance is $R = \frac{V}{I_g} - G$, where $V = 25 \, V$, $I_g = 0.01 \, A$, and $G = 50 \, \Omega$.
Substituting the values: $R = \frac{25}{0.01} - 50 = 2500 - 50 = 2450 \, \Omega$.
Therefore, a resistance of $2450 \, \Omega$ should be connected in series.

(D) Given: Galvanometer resistance $G = 50 \,\Omega$,full-scale deflection current $I_g = 100 \,\mu A = 100 \times 10^{-6} \, A = 10^{-4} \, A$,and the desired range $I = 10 \, mA = 10 \times 10^{-3} \, A = 10^{-2} \, A$.
To convert a galvanometer into an ammeter,a shunt resistance $S$ is connected in parallel with the galvanometer.
The formula for shunt resistance is $S = \frac{I_g \times G}{I - I_g}$.
Substituting the values: $S = \frac{10^{-4} \times 50}{10^{-2} - 10^{-4}} = \frac{50 \times 10^{-4}}{100 \times 10^{-4} - 1 \times 10^{-4}} = \frac{50 \times 10^{-4}}{99 \times 10^{-4}} = \frac{50}{99} \approx 0.505 \,\Omega$.
Rounding to the nearest provided option,the shunt resistance is $0.5 \,\Omega$ in parallel.

(A) voltmeter is a device used to measure the potential difference between two points in an electrical circuit.
To measure voltage,the voltmeter must be connected in parallel to the component across which the potential difference is to be measured.
To ensure that the voltmeter does not draw significant current from the circuit (which would alter the potential difference being measured),it must have a very high resistance.
This is achieved by connecting a high resistance in series with a galvanometer.
Therefore,the correct option is $A$.

(A) The galvanometer resistance $G = 100 \,\Omega$ and shunt resistance $S = 0.1 \,\Omega$.
The maximum current through the galvanometer is $I_G = 100 \,\mu A = 100 \times 10^{-6} \, A = 0.1 \, mA$.
For an ammeter,the shunt resistance $S$ is connected in parallel with the galvanometer.
The potential difference across the galvanometer and the shunt is the same: $I_G \times G = (I - I_G) \times S$.
Rearranging for the total current $I$: $I = I_G \left( 1 + \frac{G}{S} \right)$.
Substituting the values: $I = 0.1 \, mA \times \left( 1 + \frac{100}{0.1} \right)$.
$I = 0.1 \, mA \times (1 + 1000) = 0.1 \times 1001 \, mA$.
$I = 100.1 \, mA$.

(B) Given: Galvanometer resistance $G = 100\,\Omega$,full-scale current $I_g = 50\,\mu A = 50 \times 10^{-6}\,A$.
For a voltmeter (series resistance $R$):
$R = \frac{V}{I_g} - G = \frac{10}{50 \times 10^{-6}} - 100 = 200,000 - 100 = 199,900\,\Omega \approx 200\,k\Omega$.
Thus,option $(b)$ is correct.
For an ammeter (shunt resistance $S$):
$S = \frac{I_g G}{I - I_g} = \frac{50 \times 10^{-6} \times 100}{10 \times 10^{-3} - 50 \times 10^{-6}} = \frac{5 \times 10^{-3}}{10^{-2} - 0.05 \times 10^{-3}} = \frac{5 \times 10^{-3}}{9.95 \times 10^{-3}} \approx 0.5\,\Omega$.
Options $(c)$ and $(d)$ are incorrect.

(B) The voltage sensitivity $(V_s)$ is related to current sensitivity $(I_s)$ and galvanometer resistance $(G)$ by the formula: $V_s = \frac{I_s}{G}$.
Given $I_s = 10 \, \text{div/mA}$ and $V_s = 2 \, \text{div/mV}$,we find $G = \frac{I_s}{V_s} = \frac{10}{2} = 5 \, \Omega$.
The full-scale deflection current $(I_g)$ is the total divisions divided by current sensitivity: $I_g = \frac{150 \, \text{div}}{10 \, \text{div/mA}} = 15 \, \text{mA} = 15 \times 10^{-3} \, \text{A}$.
To make each division read $1 \, V$,the total voltage $(V)$ to be measured for full-scale deflection is $150 \, \text{divisions} \times 1 \, \text{V/division} = 150 \, \text{V}$.
To convert the galvanometer into a voltmeter of range $V$,a series resistance $R$ is required,given by $R = \frac{V}{I_g} - G$.
Substituting the values: $R = \frac{150}{15 \times 10^{-3}} - 5 = 10000 - 5 = 9995 \, \Omega$.

(D) To convert a galvanometer into a voltmeter,a high resistance $R$ is connected in series with the galvanometer of resistance $G$.
The total resistance of the voltmeter is $R_V = R + G$.
The full-scale deflection voltage $V$ is given by $V = i_g(R + G)$,where $i_g$ is the full-scale deflection current of the galvanometer.
Rearranging this,we get $R_V = R + G = \frac{V}{i_g}$.
Since $i_g$ is a constant for a given galvanometer,we have $R_V \propto V$.
This represents a linear relationship passing through the origin (if we consider the total resistance $R_V$ as a function of $V$).
Therefore,the graph is a straight line passing through the origin,which corresponds to graph $(iv)$.

(B) Let $I_g$ be the full-scale deflection current of the galvanometer and $G$ be its resistance. The initial range is $I_0 = I_g = 1 \, A$.
When a shunt resistance $S$ is connected in parallel,the new range $I$ is given by the formula: $I = I_g \left(1 + \frac{G}{S}\right)$.
Since $I_g = 1 \, A$,we have $I = 1 + \frac{G}{S}$.
Rearranging for $S$,we get: $S = \frac{G}{I - 1}$.
This equation represents a rectangular hyperbola where $S \to \infty$ as $I \to 1$ and $S \to 0$ as $I \to \infty$. Looking at the graph,the curve $Q$ represents this hyperbolic relationship starting from the range $1 \, A$ on the $x$-axis.

(D) . The Duddell's thermo-galvanometer is an instrument used for measuring small electric currents. It is suitable for circuits of any frequency (even up to $120,000 \, Hz$) and can measure currents as small as $20 \, \mu A$. It is equally effective for both direct current $(DC)$ and alternating current $(AC)$. Therefore,statement $A$ is false.
$B$. $A$ thermopile is a device consisting of several thermocouples connected in series,used to measure small amounts of radiant heat. It is highly sensitive and can detect temperature differences of the order of $10^{-3} \, ^\circ C$. Therefore,statement $B$ is true.
Conclusion: Statement $A$ is false,and statement $B$ is true.

(A) The actual current $i$ flowing through the circuit can be calculated using Faraday's law of electrolysis: $m = Z i t$.
Given: $m = 2.0124\, g$, $Z = 1.118 \times 10^{-3}\, g\, C^{-1}$, and $t = 1\, \text{hour} = 3600\, s$.
Rearranging for current: $i = \frac{m}{Z t}$.
Substituting the values: $i = \frac{2.0124}{1.118 \times 10^{-3} \times 3600} = \frac{2.0124}{4.0248} = 0.5\, A$.
The error in the ammeter reading is defined as: $\text{Error} = \text{Measured value} - \text{Actual value}$.
$\text{Error} = 0.54\, A - 0.5\, A = + 0.04\, A$.

(C) In a moving coil galvanometer,a radial magnetic field is essential to ensure that the torque on the coil remains constant regardless of its orientation. This is achieved by using a soft iron core and shaping the pole pieces of the magnet into a cylindrical form. This geometry ensures that the magnetic field lines are always parallel to the plane of the coil,making the field radial.

(B) The deflection $\theta$ in a moving coil galvanometer is given by the formula $\theta = \frac{NiAB}{k}$,where $N$ is the number of turns,$i$ is the current,$A$ is the area of the coil,$B$ is the magnetic field,and $k$ is the torsional constant of the spring.
From this relation,it is clear that $\theta \propto N$.
Therefore,the deflection is directly proportional to the number of turns in the coil.

(C) To control the speed of deflection and bring the coil to rest quickly,electromagnetic damping is required. This is achieved by winding the coil on a non-magnetic metallic frame (usually copper). When the coil moves,eddy currents are induced in the metallic frame,which oppose the motion of the coil,thereby providing the necessary damping. Thus,connecting a small copper wire across its terminals or using a metallic frame provides the required damping effect. Among the given options,connecting a low resistance (like a copper wire) across the terminals creates a closed circuit for eddy currents,providing damping.

(B) In a moving coil galvanometer,the magnetic torque acting on the coil is given by $\tau_m = NIAB \sin \phi$. For a radial magnetic field,$\phi = 90^\circ$,so $\tau_m = NIAB$.
This magnetic torque is balanced by the restoring torque of the spring,$\tau_r = C\theta$,where $C$ is the torsional constant.
Equating the two torques: $NIAB = C\theta$.
Rearranging for current: $i = \left( \frac{C}{NAB} \right) \theta$.
Since $C, N, A,$ and $B$ are constants for a given galvanometer,we have $i \propto \theta$.

(D) The torque $\tau$ acting on a current-carrying coil placed in a magnetic field is given by the formula $\tau = \vec{m} \times \vec{B}$,where $\vec{m}$ is the magnetic dipole moment and $\vec{B}$ is the magnetic field.
For a coil with $N$ turns,area $A$,and current $I$,the magnetic moment is $m = NIA$.
The magnitude of the torque is $\tau = mB \sin \theta$,where $\theta$ is the angle between the normal to the coil and the magnetic field.
In a moving coil galvanometer,the magnetic field is made radial so that the plane of the coil is always parallel to the magnetic field lines. This means the angle between the normal to the coil and the magnetic field is always $90^\circ$.
Therefore,$\sin 90^\circ = 1$.
Substituting these values,we get $\tau = (NIA)B \sin 90^\circ = NABI$.

(D) In a pivoted coil galvanometer, the magnetic field must be radial so that the plane of the coil is always parallel to the magnetic field lines, regardless of its orientation. This ensures that the torque on the coil remains constant $( \tau = NIAB )$ for a given current. To achieve this radial magnetic field, the pole pieces of the horse-shoe magnet are made cylindrical (concave). Thus, the correct option is $D$.

(D) The sensitivity $S$ of a moving coil galvanometer is given by the formula $S = \frac{\theta}{i} = \frac{nBA}{k}$,where $n$ is the number of turns,$B$ is the magnetic field,$A$ is the area of the coil,and $k$ (or $C$) is the couple per unit twist of the suspension.
From the formula,it is clear that the sensitivity $S$ is directly proportional to $n$,$B$,and $A$,and inversely proportional to $k$.
Therefore,to increase the sensitivity $S$,we must decrease the value of $k$ (the couple per unit twist of the suspension).

(D) The current sensitivity of a galvanometer is defined as the deflection produced per unit current,given by the formula: $\frac{\theta}{i} = \frac{NBA}{k}$.
Given:
Number of turns $N = 100$
Magnetic field $B = 5 \, T$
Area $A = 1 \, cm^2 = 10^{-4} \, m^2$
Restoring couple per unit radian $k = 10^{-8} \, N \cdot m/rad$.
Substituting the values:
$\frac{\theta}{i} = \frac{100 \times 5 \times 10^{-4}}{10^{-8}}$
$\frac{\theta}{i} = \frac{5 \times 10^{-2}}{10^{-8}} = 5 \times 10^6 \, rad/A$.
Since $1 \, A = 10^6 \, \mu A$,we have:
$\frac{\theta}{i} = 5 \times 10^6 \times 10^{-6} \, rad/\mu A = 5 \, rad/\mu A$.

(D) The current sensitivity $(I_s)$ of a moving coil galvanometer is defined as the deflection produced per unit current,given by the formula:
$I_s = \frac{\theta}{i} = \frac{NAB}{k}$
Where:
$N$ = Number of turns in the coil
$A$ = Area of the coil
$B$ = Magnetic field strength
$k$ = Torsional constant of the spring
From the formula,it is evident that $I_s$ is directly proportional to $N$,$A$,and $B$.
Therefore,increasing the magnetic field $(B)$,the area of the coil $(A)$,or the number of turns $(N)$ will increase the current sensitivity.
Hence,the correct option is $(d)$.

(B) The current sensitivity of a moving coil galvanometer is given by the formula $S = \frac{NAB}{k}$,where $N$ is the number of turns,$A$ is the area of the coil,$B$ is the magnetic field strength,and $k$ (or $C$) is the torsional constant of the suspension wire.
To increase the sensitivity $S$,we must increase $N$,$A$,or $B$,or decrease the torsional constant $k$.
Therefore,decreasing the torsional constant of the suspension is the correct method to increase sensitivity.
Thus,the correct option is $B$.

(A) Current sensitivity $({\sigma _i})$ is defined as the deflection per unit current: ${\sigma _i} = \frac{\theta}{I}$.
Voltage sensitivity $({\sigma _V})$ is defined as the deflection per unit voltage: ${\sigma _V} = \frac{\theta}{V}$.
Since $V = I \cdot G$,where $G$ is the resistance of the galvanometer,we can substitute $V$ in the voltage sensitivity equation:
${\sigma _V} = \frac{\theta}{I \cdot G}$.
Substituting ${\sigma _i} = \frac{\theta}{I}$ into the expression for ${\sigma _V}$,we get:
${\sigma _V} = \frac{{\sigma _i}}{G}$.
Therefore,the relation is $\frac{{\sigma _i}}{G} = {\sigma _V}$.

(A) In a moving coil galvanometer,a radial magnetic field is essential to ensure that the magnetic torque remains constant regardless of the coil's orientation. This is achieved by using a strong permanent magnet with concave-shaped pole pieces. These concave pole pieces,along with a cylindrical soft iron core,ensure that the magnetic field lines are always parallel to the plane of the coil,making the field radial. Therefore,the correct shape is concave.

(A) soft iron core is used in a moving coil galvanometer. The soft iron core attracts the magnetic lines of force,which increases the strength of the magnetic field.
This increases the sensitivity of the galvanometer.
Furthermore,the use of a cylindrical soft iron core makes the magnetic field radial,ensuring that the plane of the coil remains parallel to the direction of the magnetic field for all orientations of the coil.

(A) Sensitivity $(S)$ is defined as the deflection per unit current,given by $S = \frac{\theta}{i}$.
For the same deflection $\theta = 10$ divisions:
$S_A = \frac{10}{3\,mA}$ and $S_B = \frac{10}{5\,mA}$.
Taking the ratio: $\frac{S_A}{S_B} = \frac{10/3}{10/5} = \frac{5}{3}$.
Since $\frac{S_A}{S_B} = 1.67 > 1$,it follows that $S_A > S_B$.
Therefore,galvanometer $A$ is more sensitive than $B$.

(B) In a moving coil galvanometer,the torque acting on the coil is given by $\tau = NIAB \sin \theta$. In the equilibrium position,this magnetic torque is balanced by the restoring torque of the spring,$\tau_r = k \phi$,where $\phi$ is the deflection. Thus,$NIAB = k \phi$,which implies $\phi = (\frac{NAB}{k}) i$. Since $N, A, B,$ and $k$ are constants,we have $\phi \propto i$. Therefore,if the current $i$ is doubled,the deflection $\phi$ is also doubled. In a tangent galvanometer,the current is proportional to the tangent of the deflection $(i \propto \tan \theta)$,so the deflection does not double when the current doubles.

(B) In a moving coil galvanometer,the coil is wound on a metallic (copper or aluminum) frame to make the motion dead-beat due to the production of eddy currents,which provide electromagnetic damping.
In a ballistic galvanometer,the objective is to measure the total charge passing through the circuit,which requires the coil to oscillate with minimal damping.
Therefore,the frame is made of a non-conducting (non-metallic) material,such as paper or bamboo,to avoid the production of eddy currents that would otherwise damp the motion of the coil.

(B) The sensitivity $S$ of a tangent galvanometer is defined as the deflection produced per unit current,given by $S = \frac{\theta}{i}$.
From the principle of the tangent galvanometer,$i = K \tan \theta$,where $K = \frac{2R B_H}{\mu_0 N}$ is the reduction factor.
Substituting $i$ in the sensitivity formula,we get $S = \frac{\theta}{K \tan \theta} = \frac{1}{K} \cdot \frac{\theta}{\tan \theta}$.
For small angles,$\tan \theta \approx \theta$,so $S \approx \frac{1}{K} = \frac{\mu_0 N}{2R B_H}$.
From this expression,it is clear that the sensitivity $S$ is directly proportional to the number of turns $N$.
Therefore,to increase the sensitivity,the number of turns $N$ must be increased.

(B) The $Moving$ $coil$ $galvanometer$,$Dynamo$,and $Electric$ $motor$ all operate based on the magnetic effects of current or electromagnetic induction.
$A$ $Hot$ $wire$ $ammeter$ works on the principle of the heating effect of current $(H = I^2Rt)$,where the current passing through a wire causes it to heat up and expand,leading to a deflection of the pointer.
Therefore,it does not depend on the magnetic effect of current.

(A) dead-beat galvanometer is designed to bring the pointer to rest quickly without oscillations.
When the coil of the galvanometer moves in the magnetic field,it induces eddy currents in the metallic frame (usually made of aluminum or copper) on which the coil is wound.
According to Lenz's Law,these eddy currents oppose the motion of the coil,providing magnetic damping.
This damping effect causes the pointer to come to a steady deflection rapidly,making the galvanometer 'dead-beat'.

(C) An ammeter is used to measure the current flowing through a circuit. To measure the total current,it must be connected in series with the conductor so that the entire current passes through it. Connecting it in parallel would cause a short circuit due to its very low resistance. Therefore,the statement that an ammeter is connected in parallel is incorrect.

(B) voltmeter is connected in parallel to the component across which the potential difference is to be measured. If the resistance of the voltmeter is low,it would draw a significant amount of current from the circuit,thereby altering the potential difference it is intended to measure. By keeping the resistance of the voltmeter very high,it draws negligible current from the circuit,ensuring that the potential difference across the component remains practically unchanged.

(B) Let $I_g$ be the full-scale deflection current of the galvanometer and $G$ be its resistance.
The initial range of the ammeter is $I_0 = I_g = 1 \ A$.
When a shunt resistance $S$ is connected in parallel to increase the range to $I$,the formula is given by $S = \frac{I_g G}{I - I_g}$.
Substituting $I_g = 1 \ A$,we get $S = \frac{G}{I - 1}$.
This equation represents a rectangular hyperbola of the form $y = \frac{k}{x - c}$,where $S$ is the shunt resistance and $I$ is the range.
As $I$ increases,$S$ decreases,and as $I \to 1$,$S \to \infty$.
Looking at the provided graph,curve $Q$ represents this hyperbolic relationship where the shunt resistance decreases as the range increases,starting from infinity at $I = 1 \ A$.

(D) Let the initial current be $I$ and the resistance of the coil be $G = 90\, \Omega$.
We want the current through the coil $(I_g)$ to be reduced by $90\%$,which means the new current $I_g = I - 0.9I = 0.1I = I/10$.
Using the formula for shunt resistance $S$ required to divert current such that the galvanometer current becomes $I_g = I/n$,where $n = I/I_g = 10$.
The shunt resistance is given by $S = \frac{G}{n - 1}$.
Substituting the values: $S = \frac{90}{10 - 1} = \frac{90}{9} = 10\, \Omega$.
Therefore,a resistor of $10\, \Omega$ must be connected in parallel.

(D) Given: Resistance of galvanometer $R_g = 990 \,\Omega$,full-scale deflection current $I_g = 10 \,mA = 0.01 \,A$,and the desired range of the ammeter $I = 1 \,A$.
To convert a galvanometer into an ammeter,a shunt resistance $S$ is connected in parallel with it.
The formula for shunt resistance is $S = \frac{I_g R_g}{I - I_g}$.
Substituting the values: $S = \frac{0.01 \times 990}{1 - 0.01}$.
$S = \frac{9.9}{0.99} = 10 \,\Omega$.
Therefore,the required shunt resistance is $10 \,\Omega$.

(C) The current required for full-scale deflection is $I_g = 20 \, \mu A/\text{division} \times 30 \, \text{divisions} = 600 \, \mu A = 6 \times 10^{-4} \, A$.
To convert a galvanometer into a voltmeter,a high resistance $R$ must be connected in series with it.
The formula for series resistance is $R = \frac{V}{I_g} - R_g$.
Substituting the given values: $R = \frac{1}{6 \times 10^{-4}} - 25$.
$R = 1666.67 - 25 = 1641.67 \, \Omega$.

(A) Let the original resistance of the galvanometer be $G$. When a shunt resistance $S$ is connected in parallel,the equivalent resistance of the combination is $R_{eq} = \frac{GS}{G + S}$.
To keep the main current unchanged,the total resistance of the circuit must remain equal to the original resistance $G$.
Let $S'$ be the resistance connected in series with the galvanometer. The new equivalent resistance is $R'_{eq} = \frac{GS}{G + S} + S'$.
Setting $R'_{eq} = G$,we get:
$G = \frac{GS}{G + S} + S'$
$S' = G - \frac{GS}{G + S}$
$S' = \frac{G(G + S) - GS}{G + S}$
$S' = \frac{G^2 + GS - GS}{G + S}$
$S' = \frac{G^2}{G + S}$

(B) The current required for full-scale deflection is $I_g = 20 \, \mu A/division \times 30 \, divisions = 600 \, \mu A = 6 \times 10^{-4} \, A$.
To convert the galvanometer into an ammeter,a shunt resistance $S$ must be connected in parallel with it.
The formula for shunt resistance is $S = \frac{I_g R_g}{I - I_g}$.
Given $I_g = 6 \times 10^{-4} \, A$,$R_g = 25 \, \Omega$,and $I = 1 \, A$.
$S = \frac{(6 \times 10^{-4}) \times 25}{1 - 6 \times 10^{-4}} = \frac{0.015}{0.9994} \approx 0.015 \, \Omega$.
Therefore,a shunt of $0.015 \, \Omega$ should be connected in parallel.

(B) Let the resistance $R$ be connected in series with the voltmeter.
The full-scale deflection current of the voltmeter is $i_g = V/G$.
When the range is increased to $nV$,the total resistance becomes $G + R$.
According to Ohm's law,the new voltage range is given by:
$nV = i_g(G + R)$
Substituting $i_g = V/G$:
$nV = (V/G)(G + R)$
$n = (G + R)/G$
$nG = G + R$
$R = nG - G = (n - 1)G$

(B) The resistance $R$ is given by the ratio of the potential difference across the resistor to the current flowing through it.
$R = \frac{V_R}{I_R}$
In the given circuit,the voltmeter is connected in parallel with the resistor $R$. Therefore,the potential difference across the resistor is $20 \ V$.
However,the ammeter is connected in series with the parallel combination of the voltmeter and the resistor. The total current $I = 4 \ A$ splits into two parts: a current $I_V$ flowing through the voltmeter and a current $I_R$ flowing through the resistor.
Thus,$I_R = I - I_V = 4 \ A - I_V$.
Since $I_V > 0$,the current flowing through the resistor $I_R$ must be less than $4 \ A$.
Therefore,$R = \frac{20 \ V}{I_R} > \frac{20 \ V}{4 \ A} = 5 \ \Omega$.
Hence,the resistance $R$ is greater than $5 \ \Omega$.

(D) Let the initial current through the galvanometer be $I_g = I/5$. The shunt resistance is $S = 4\,\Omega$. The current through the shunt is $I_s = I - I_g = I - I/5 = 4I/5$.
Since the galvanometer and shunt are in parallel,the potential difference across them is equal:
$I_g \cdot G = I_s \cdot S$
$(I/5) \cdot G = (4I/5) \cdot 4$
$G = 16\,\Omega$
Now,an additional shunt of $2\,\Omega$ is connected in parallel to the existing $4\,\Omega$ shunt. The new equivalent shunt resistance $S'$ is:
$1/S' = 1/4 + 1/2 = (1+2)/4 = 3/4$
$S' = 4/3\,\Omega$
Let the new current through the galvanometer be $I'$. The total current $I$ remains the same. The potential difference across the galvanometer and the new shunt is equal:
$I' \cdot G = (I - I') \cdot S'$
$I' \cdot 16 = (I - I') \cdot (4/3)$
$12I' = I - I'$
$13I' = I$
$I' = I/13$

(B) The full-scale deflection current $I_g$ is calculated as:
$I_g = (\text{number of divisions}) \times (\text{current sensitivity})$
$I_g = 25 \times 4 \times 10^{-4} \,A = 10^{-2} \,A$
To convert a galvanometer into a voltmeter, a high resistance $R$ is connected in series with it.
The formula for the series resistance is:
$R = \frac{V}{I_g} - G$
where $V = 2.5 \,V$ and $G = 100 \,\Omega$.
$R = \frac{2.5}{10^{-2}} - 100$
$R = 250 - 100 = 150 \,\Omega$
Therefore, a resistance of $150 \,\Omega$ must be connected in series.

(D) To convert an ammeter into a voltmeter, a high resistance $R$ is connected in series with the ammeter.
The formula for the series resistance is $R = \frac{V}{I_g} - G$, where $V$ is the desired voltage range, $I_g$ is the full-scale deflection current of the ammeter, and $G$ is the internal resistance of the ammeter.
Given: $V = 5 \,V$, $I_g = 100 \,mA = 100 \times 10^{-3} \,A = 0.1 \,A$, and $G = 2 \,\Omega$.
Substituting the values: $R = \frac{5}{0.1} - 2$.
$R = 50 - 2 = 48 \,\Omega$.
Thus, a resistance of $48 \,\Omega$ must be connected in series.

(A) To convert an ammeter into a voltmeter,a high resistance $R$ must be connected in series with the ammeter.
Given:
Resistance of ammeter $G = 1\, \Omega$
Full-scale current $I_g = 10\, mA = 10 \times 10^{-3}\, A = 0.01\, A$
Required voltage range $V = 10\, V$
The formula for series resistance is $R = \frac{V}{I_g} - G$.
Substituting the values:
$R = \frac{10}{0.01} - 1$
$R = 1000 - 1$
$R = 999\, \Omega$.
Therefore,a resistance of $999\, \Omega$ must be connected in series.

(C) Let the resistance of the ammeter be $G = 0.018 \, \Omega$ and the full-scale current be $I_g = 1 \, A$.
We want to measure a total current $I = 10 \, A$ by connecting a shunt resistance $S$ in parallel.
The formula for the shunt resistance is $S = \frac{I_g G}{I - I_g}$.
Substituting the given values: $S = \frac{1 \times 0.018}{10 - 1}$.
$S = \frac{0.018}{9} = 0.002 \, \Omega$.
Therefore,a shunt resistance of $0.002 \, \Omega$ is required.