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

(C) Given,resistance of galvanometer $R_g = 100 \Omega$.
Full scale deflection current $I_g = 50 \mu A = 5 \times 10^{-5} \text{ A}$.
Required range of ammeter $I = 10 \text{ mA} = 10^{-2} \text{ A}$.
To convert a galvanometer into an ammeter,a shunt resistance $R_s$ is connected in parallel.
The formula for shunt resistance is $R_s = \frac{I_g}{I - I_g} \times R_g$.
Substituting the values:
$R_s = \frac{5 \times 10^{-5}}{10^{-2} - 5 \times 10^{-5}} \times 100$
$R_s = \frac{5 \times 10^{-5}}{100 \times 10^{-4} - 0.5 \times 10^{-4}} \times 100$
$R_s = \frac{5 \times 10^{-5}}{995 \times 10^{-5}} \times 100$
$R_s = \frac{500}{995} \approx 0.5025 \Omega$.
Thus,the resistance to be added is approximately $0.5 \Omega$.

(A) Given: Galvanometer resistance $R_G = 40 \Omega$, Shunt resistance $R_S = 2 \Omega$, Sensitivity $= 10 \text{ div/mA}$, Total divisions $n = 50$.
The maximum current that the galvanometer can measure without shunt is $I_G = \frac{50 \text{ div}}{10 \text{ div/mA}} = 5 \text{ mA}$.
When a shunt $R_S$ is connected in parallel, the total current $I$ is divided between the galvanometer and the shunt.
Using the principle of parallel circuits, the voltage across the galvanometer and shunt is equal: $I_G R_G = I_S R_S$.
Since $I = I_G + I_S$, we have $I_S = I - I_G$.
Substituting this: $I_G R_G = (I - I_G) R_S$.
Rearranging for $I$: $I = I_G \left( 1 + \frac{R_G}{R_S} \right) = I_G \left( \frac{R_G + R_S}{R_S} \right)$.
Substituting the values: $I = 5 \text{ mA} \times \left( \frac{40 + 2}{2} \right) = 5 \times \frac{42}{2} = 5 \times 21 = 105 \text{ mA}$.

(B) When a voltmeter of resistance $R$ is connected in parallel with a resistance $r$,the equivalent resistance of the combination is $R_{eq} = \frac{Rr}{R+r}$.
The potential difference measured by the voltmeter is $V' = I \times R_{eq} = I \times \frac{Rr}{R+r}$,where $I$ is the current through the branch.
The actual potential difference across $r$ in the absence of the voltmeter is $V = I \times r$.
To make the measured potential difference $V'$ as close as possible to the actual potential difference $V$,we need the ratio $\frac{V'}{V} = \frac{R}{R+r}$ to be as close to $1$ as possible.
This occurs when $R \gg r$ (i.e.,$R$ is much greater than $r$).
Therefore,the reading is nearest to the actual value when $R > r$.

(D) Case $I$: Shunt resistance $R_1$ for a galvanometer $G$ to pass $\frac{1}{n}$ of the main current $i$ is given by $R_1 = \frac{G}{n-1}$.
Here,$n = 51$,so $R_1 = \frac{G}{51-1} = \frac{G}{50}$.
Case $II$: Series resistance $R_2$ for a galvanometer $G$ to measure $\frac{1}{m}$ of the main voltage $V$ is given by $R_2 = G(m-1)$.
Here,$m = 11$,so $R_2 = G(11-1) = 10G$.
Now,calculating the ratio $\frac{R_2}{R_1}$:
$\frac{R_2}{R_1} = \frac{10G}{G/50} = 10G \times \frac{50}{G} = 500$.

(A) Given,the ratio of heats generated through the shunt $(H_s)$ and galvanometer $(H_g)$ is $H_s : H_g = 7 : 5$.
The resistance of the galvanometer is $R_g = 112 \Omega$.
Since the shunt and galvanometer are connected in parallel to form an ammeter,the potential difference $(V)$ across both is the same.
The heat generated in a resistor is given by $H = \frac{V^2}{R} \times t$.
Therefore,the ratio of heat generated is $\frac{H_s}{H_g} = \frac{V^2 / R_s}{V^2 / R_g} = \frac{R_g}{R_s}$.
Given $\frac{H_s}{H_g} = \frac{7}{5}$,we have $\frac{R_g}{R_s} = \frac{7}{5}$.
Substituting the value of $R_g = 112 \Omega$:
$\frac{112}{R_s} = \frac{7}{5}$
$R_s = \frac{112 \times 5}{7}$
$R_s = 16 \times 5 = 80 \Omega$.
Thus,the resistance of the shunt is $80 \Omega$.

(A) Let the initial resistance of the galvanometer be $G$. When a shunt resistance $S$ is connected in parallel with the galvanometer,the equivalent resistance of the combination is $R_{eq} = \frac{G \times S}{G+S}$.
To keep the main current unchanged,the total resistance of the circuit must remain equal to the initial resistance $G$. Let a resistance $R$ be connected in series with this parallel combination.
Therefore,the total resistance is $R_{total} = R + \frac{G \times S}{G+S}$.
Setting $R_{total} = G$,we get:
$G = R + \frac{GS}{G+S}$
$R = G - \frac{GS}{G+S}$
$R = \frac{G(G+S) - GS}{G+S}$
$R = \frac{G^2 + GS - GS}{G+S}$
$R = \frac{G^2}{G+S}$
Thus,the required series resistance is $\frac{G^2}{S+G}$.
Hence,the correct option is $A$.

(A) The galvanometer resistance is $G = 10 \Omega$. The full-scale deflection current is $I_g = 1 \text{ mA} = 1 \times 10^{-3} \text{ A}$. The new range is $I = 101 \text{ mA} = 101 \times 10^{-3} \text{ A}$.
To convert a galvanometer into an ammeter,a shunt resistance $S$ is connected in parallel.
The formula for shunt resistance is $S = \frac{I_g G}{I - I_g}$.
Substituting the values: $S = \frac{1 \times 10^{-3} \times 10}{101 \times 10^{-3} - 1 \times 10^{-3}} = \frac{10 \times 10^{-3}}{100 \times 10^{-3}} = 0.1 \Omega$.
The resistance of a wire is given by $R = \rho \frac{L}{A}$,where $\rho = 1 \times 10^{-6} \Omega \text{ m}$ and $A = 1 \text{ mm}^2 = 1 \times 10^{-6} \text{ m}^2$.
Setting $R = S = 0.1 \Omega$,we get $0.1 = (1 \times 10^{-6}) \times \frac{L}{1 \times 10^{-6}}$.
$0.1 = L$.
Thus,$L = 0.1 \text{ m} = 10 \text{ cm}$.

(A) The current sensitivity $(I_s)$ of a moving coil galvanometer is given by the formula: $I_s = \frac{NBA}{k}$,where $N$ is the number of turns,$B$ is the magnetic field,$A$ is the area of the coil,and $k$ is the torsional constant of the spring.
Since the springs are identical,$k_A = k_B = k$.
The ratio of current sensitivities is $\frac{I_{sA}}{I_{sB}} = \frac{N_A B_A A_A}{N_B B_B A_B}$.
Given values:
$N_A = 36, B_A = 0.25 \ T, A_A = 2.4 \times 10^{-3} \ m^2$
$N_B = 48, B_B = 0.5 \ T, A_B = 4.8 \times 10^{-3} \ m^2$
Substituting these values:
$\frac{I_{sA}}{I_{sB}} = \frac{36 \times 0.25 \times 2.4 \times 10^{-3}}{48 \times 0.5 \times 4.8 \times 10^{-3}}$
$\frac{I_{sA}}{I_{sB}} = \left(\frac{36}{48}\right) \times \left(\frac{0.25}{0.5}\right) \times \left(\frac{2.4}{4.8}\right)$
$\frac{I_{sA}}{I_{sB}} = \left(\frac{3}{4}\right) \times \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) = \frac{3}{16}$.
Thus,the ratio is $3: 16$.

(B) Current sensitivity is defined as the deflection produced per unit current flowing through the galvanometer.
It is given by the formula: $I_{S} = \frac{\theta}{I}$
Where $\theta$ is the deflection in divisions and $I$ is the current in amperes.
Given: $\theta = 25 \ div$ and $I = 0.1 \ A$.
Therefore,$I_{S} = \frac{25}{0.1} = 250 \ div/A$.

(B) Given that, for a moving coil galvanometer:
Effective area, $A = 4 \times 10^{-2} \, m^2$
Magnetic field, $B = 5 \times 10^{-2} \, Wb \, m^{-2}$
Angle of deflection, $\phi = 0.2 \, rad$
Electric current, $I = 5 \, mA = 5 \times 10^{-3} \, A$
Assuming the number of turns $N = 1$ (as not specified, we calculate the product $NAB$):
The torque balance equation is $NAB I = k \phi$, where $k$ is the torsional constant.
$k = \frac{NAB I}{\phi} = \frac{(1) \times (4 \times 10^{-2}) \times (5 \times 10^{-2}) \times (5 \times 10^{-3})}{0.2}$
$k = \frac{100 \times 10^{-7}}{0.2} = 500 \times 10^{-7} = 5 \times 10^{-5} \, N \, m \, rad^{-1}$.
Current sensitivity $S_I = \frac{\phi}{I} = \frac{0.2}{5 \times 10^{-3}} = 40 \, rad \, A^{-1}$.
Thus, the current sensitivity is $40 \, rad \, A^{-1}$.

(A) Current sensitivity $I_S$ is given by $I_S = \frac{NBA}{k}$. Assuming the spring constant $k$ is the same for both galvanometers,the ratio of current sensitivities is:
$\frac{I_{SX}}{I_{SY}} = \frac{N_X B_X A_X}{N_Y B_Y A_Y} = \frac{30 \times 0.25 \times 4.8 \times 10^{-3}}{45 \times 0.50 \times 2.4 \times 10^{-3}} = \frac{30}{45} \times \frac{0.25}{0.50} \times \frac{4.8}{2.4} = \frac{2}{3} \times \frac{1}{2} \times 2 = \frac{2}{3}$.
Voltage sensitivity $V_S$ is given by $V_S = \frac{I_S}{R} = \frac{NBA}{kR}$. The ratio of voltage sensitivities is:
$\frac{V_{SX}}{V_{SY}} = \frac{I_{SX}}{I_{SY}} \times \frac{R_Y}{R_X} = \frac{2}{3} \times \frac{14}{10} = \frac{2}{3} \times \frac{7}{5} = \frac{14}{15}$.
Thus,the ratios are $2 : 3$ and $14 : 15$.

(A) The voltage sensitivity $(V_s)$ of a moving coil galvanometer is given by $V_s = \frac{NBA}{kR}$,where $N$ is the number of turns,$B$ is the magnetic field,$A$ is the area,$k$ is the torsional constant,and $R$ is the resistance.
Given the ratio of voltage sensitivities: $\frac{V_{sA}}{V_{sB}} = \frac{4}{3}$.
Given the ratio of resistances: $\frac{R_A}{R_B} = \frac{3}{4}$.
Given the ratio of areas: $\frac{A_A}{A_B} = \frac{1}{2}$.
Since $B$ and $k$ are constant,we have $\frac{V_{sA}}{V_{sB}} = \frac{N_A A_A R_B}{N_B A_B R_A}$.
Substituting the given values: $\frac{4}{3} = \frac{N_A}{N_B} \times (\frac{1}{2}) \times (\frac{4}{3})$.
$\frac{4}{3} = \frac{N_A}{N_B} \times \frac{2}{3}$.
$\frac{N_A}{N_B} = \frac{4}{3} \times \frac{3}{2} = 2$.
Given $N_A = 200$,then $N_B = \frac{N_A}{2} = \frac{200}{2} = 100$.

(B) Let the total current in the circuit be $I$.
The current passing through the galvanometer is $I_G = 0.05 I$.
The current passing through the shunt resistance $S$ is $I_S = I - I_G = I - 0.05 I = 0.95 I$.
Since the galvanometer and the shunt resistance are in parallel,the potential difference across them is equal:
$I_G G = I_S S$
Substituting the values:
$0.05 I \cdot G = 0.95 I \cdot S$
$S = \frac{0.05 I \cdot G}{0.95 I} = \frac{5}{95} G = \frac{1}{19} G$
Therefore,the shunt resistance is $S = \frac{G}{19}$.

(C) Given: Range of voltmeter $V_g = 250 mV = 0.25 V$,Resistance $G = 10 \Omega$.
First,calculate the full-scale deflection current $I_g$ of the galvanometer:
$I_g = \frac{V_g}{G} = \frac{0.25 V}{10 \Omega} = 0.025 A = 25 mA$.
We want to convert this into an ammeter of range $I = 250 mA = 0.25 A$.
The formula for the shunt resistance $S$ is:
$S = \frac{G I_g}{I - I_g}$
Substituting the values:
$S = \frac{10 \times 0.025}{0.25 - 0.025} = \frac{0.25}{0.225} = \frac{250}{225} \approx 1.11 \Omega$.
Rounding to the nearest given option,the value is $1 \Omega$.

(D) The sensitivity of a galvanometer is defined by the ratio of the current through the galvanometer $(i_g)$ to the total current $(i)$.
Given that the sensitivity is decreased by $\frac{1}{40}$ times,we have $\frac{i_g}{i} = \frac{1}{40}$.
The formula for current division in a galvanometer with a shunt resistance $(S)$ is given by:
$\frac{i_g}{i} = \frac{S}{S+G}$
where $G$ is the resistance of the galvanometer.
Substituting the given values ($S = 10 \Omega$ and $\frac{i_g}{i} = \frac{1}{40}$):
$\frac{1}{40} = \frac{10}{10+G}$
$10 + G = 400$
$G = 400 - 10 = 390 \Omega$
Therefore,the resistance of the galvanometer is $390 \Omega$.

(A) The sensitivity of a galvanometer is defined as the deflection per unit current,given by $S_g = \frac{\theta}{i_g}$. When a shunt $S$ is connected in parallel with a galvanometer of resistance $G$,the new sensitivity $S'$ is given by the ratio of the current through the galvanometer $i_g$ to the total current $i$.
The current through the galvanometer is $i_g = i \left( \frac{S}{G+S} \right)$.
Therefore,the new sensitivity is $S' = \frac{i_g}{i} = \frac{S}{G+S}$.
Given,initial sensitivity $= 60 \text{ div/A}$ and final sensitivity $= 10 \text{ div/A}$.
The ratio of sensitivities is $\frac{S'}{S_g} = \frac{10}{60} = \frac{1}{6}$.
Substituting the formula: $\frac{1}{6} = \frac{S}{G+S}$.
Cross-multiplying gives $G + S = 6S$,which simplifies to $G = 5S$.
Given $G = 20 \ \Omega$,we have $20 = 5S$.
Thus,$S = \frac{20}{5} = 4 \ \Omega$.

(C) Resistance of galvanometer,$G = 50 \Omega$.
Full scale current,$i_g = 0.05 \text{ A}$.
Maximum current to be measured,$i = 5 \text{ A}$.
Area of cross-section,$A = 2.97 \times 10^{-2} \text{ cm}^2 = 2.97 \times 10^{-6} \text{ m}^2$.
Specific resistance,$\rho = 5 \times 10^{-7} \Omega\text{-m}$.
To convert a galvanometer into an ammeter,a shunt resistance $S$ is connected in parallel.
$S = \frac{i_g G}{i - i_g} = \frac{0.05 \times 50}{5 - 0.05} = \frac{2.5}{4.95} = \frac{250}{495} = \frac{50}{99} \Omega$.
Using the formula for resistance,$S = \rho \frac{l}{A}$,we get $l = \frac{S \cdot A}{\rho}$.
$l = \frac{50}{99} \times \frac{2.97 \times 10^{-6}}{5 \times 10^{-7}} = \frac{50}{99} \times \frac{29.7}{5} = 10 \times 0.3 = 3 \text{ m}$.

(C) Given:
Resistance of the galvanometer,$G = 100 \ \Omega$.
Full-scale deflection current of the galvanometer,$I_g = 100 \ \mu A = 100 \times 10^{-6} \ A = 0.1 \times 10^{-3} \ A$.
Total current to be measured,$I = 1 \ mA = 1 \times 10^{-3} \ A$.
To convert a galvanometer into an ammeter,a shunt resistance $S$ is connected in parallel with the galvanometer.
The formula for shunt resistance is:
$S = \frac{I_g G}{I - I_g}$
Substituting the values:
$S = \frac{0.1 \times 10^{-3} \times 100}{1 \times 10^{-3} - 0.1 \times 10^{-3}}$
$S = \frac{0.1 \times 10^{-1}}{0.9 \times 10^{-3}}$
$S = \frac{10^{-2}}{0.9 \times 10^{-3}} = \frac{10}{0.9} = \frac{100}{9} \ \Omega$.

(A) Given:
Resistance of galvanometer,$G = 100 \Omega$
Shunt resistance,$S = 0.1 \Omega$
Maximum deflection current in galvanometer,$I_g = 100 \mu A = 100 \times 10^{-6} A = 10^{-4} A$
For an ammeter,the galvanometer and shunt resistance are connected in parallel. The potential difference across both is the same:
$V_{AB} = I_g G = (I - I_g) S$
Substituting the values:
$(100 \times 10^{-6}) \times 100 = (I - 100 \times 10^{-6}) \times 0.1$
$10^{-2} = (I - 10^{-4}) \times 10^{-1}$
$0.1 = I - 0.0001$
$I = 0.1 + 0.0001 = 0.1001 A$
Converting to milliamperes:
$I = 0.1001 \times 1000 mA = 100.1 mA$
Thus,the total current in the circuit is $100.1 mA$.

(A) The angular deflection $\theta$ of a moving coil galvanometer is given by the formula: $\theta = \frac{NBAi}{k}$.
Here,$N = 500$ (number of turns),$B = 0.2 \ T$ (magnetic field),$A = 0.001 \ m^2$ (area),$i = 6 \pi \times 10^{-8} \ A$ (current),and $k = 6 \times 10^{-7} \ N-m/rad$ (torsional constant).
Substituting these values into the formula:
$\theta = \frac{500 \times 0.2 \times 0.001}{6 \times 10^{-7}} \times 6 \pi \times 10^{-8}$
$\theta = \frac{100 \times 0.001}{6 \times 10^{-7}} \times 6 \pi \times 10^{-8}$
$\theta = \frac{0.1}{6 \times 10^{-7}} \times 6 \pi \times 10^{-8}$
$\theta = 0.1 \times \pi \times 10^{-1} = 0.01 \pi = \frac{\pi}{100} \ rad$.

(B) To convert a galvanometer into a voltmeter,a high resistance $R_s$ must be connected in series with the galvanometer coil.
Given:
Galvanometer resistance $R_G = 10 \Omega$
Full scale deflection current $I_g = 2 \text{ mA} = 2 \times 10^{-3} \text{ A}$
Required voltage range $V = 18 \text{ V}$
The formula for the series resistance is $V = I_g(R_G + R_s)$.
Substituting the values:
$18 = 2 \times 10^{-3} \times (10 + R_s)$
$18 / (2 \times 10^{-3}) = 10 + R_s$
$9000 = 10 + R_s$
$R_s = 9000 - 10 = 8990 \Omega$
Thus,a resistance of $8990 \Omega$ must be connected in series.

(D) Statement $(A)$ is false. Duddell's thermo galvanometer is a sensitive instrument that can measure both direct current $(DC)$ and alternating current $(AC)$ by utilizing the heating effect of the current.
Statement $(B)$ is true. $A$ thermopile consists of several thermocouples connected in series,which increases the sensitivity,allowing it to detect very small temperature differences,typically on the order of $10^{-3} {}^{\circ}C$.

(A) Let $G$ be the resistance of the galvanometer and $I_{g}$ be the current at full-scale deflection.
For the voltmeter conversion:
$V_{0} = I_{g}(G + R_{1})$
$G + R_{1} = \frac{V_{0}}{I_{g}}$
$G = \frac{V_{0}}{I_{g}} - R_{1} \dots (1)$
For the ammeter conversion:
$I_{g} G = (I_{0} - I_{g}) R_{2}$
$G = \frac{(I_{0} - I_{g}) R_{2}}{I_{g}} \dots (2)$
Equating $(1)$ and $(2)$:
$\frac{V_{0}}{I_{g}} - R_{1} = \frac{I_{0} R_{2}}{I_{g}} - R_{2}$
$\frac{V_{0} - I_{0} R_{2}}{I_{g}} = R_{1} - R_{2}$
$I_{g} = \frac{V_{0} - I_{0} R_{2}}{R_{1} - R_{2}}$

(B) The torque $\tau$ experienced by a current-carrying coil in a magnetic field is given by $\tau = N I A B \sin \theta$,where $N$ is the number of turns,$I$ is the current,$A$ is the area of the coil,$B$ is the magnetic field,and $\theta$ is the angle between the normal to the coil and the magnetic field.
Since the current $I$,magnetic field $B$,and number of turns $N$ are the same for both galvanometers,the ratio of the torques depends only on the ratio of the areas of the coils.
For the first coil (square of side $a$): $A_1 = a^2$.
For the second coil (circular of radius $r = \frac{a}{\sqrt{\pi}}$): $A_2 = \pi r^2 = \pi \left( \frac{a}{\sqrt{\pi}} \right)^2 = \pi \left( \frac{a^2}{\pi} \right) = a^2$.
Since $A_1 = A_2$,the torque experienced by both coils is the same.
Therefore,the ratio of the torque experienced by the first coil to that experienced by the second one is $1: 1$.

(A) To convert a galvanometer into a voltmeter,a high resistance $R$ must be connected in series with the galvanometer.
Given:
Internal resistance of galvanometer,$G = 10 \ \Omega$
Full-scale deflection current,$i_g = 0.01 \ A$
Required voltage range,$V = 120 \ V$
The formula for the series resistance is $R = \frac{V}{i_g} - G$.
Substituting the values:
$R = \frac{120}{0.01} - 10$
$R = 12000 - 10$
$R = 11990 \ \Omega$
Thus,a resistance of $11990 \ \Omega$ must be connected in series.

(A) Given:
Resistance of galvanometer,$G = 100 \ \Omega$
Full scale deflection current of galvanometer,$i_g = 1 \ mA$
Desired full scale deflection current of ammeter,$i = 5 \ mA$
To convert a galvanometer into an ammeter,a shunt resistance $r_s$ is connected in parallel with the galvanometer.
The formula for shunt resistance is:
$r_s = \frac{G \cdot i_g}{i - i_g}$
Substituting the given values:
$r_s = \frac{100 \times 1 \ mA}{5 \ mA - 1 \ mA}$
$r_s = \frac{100}{4} \ \Omega = 25 \ \Omega$
Thus,the required shunt resistance is $25 \ \Omega$.

(C) Current sensitivity $I_s$ is defined as the deflection per unit current,given by $I_s = \frac{\theta}{I} = \frac{NAB}{k}$.
Here,$N$ is the number of turns (dimensionless),$A$ is the area $[L^2]$,$B$ is the magnetic field,and $k$ is the restoring torque per unit twist.
The dimensions of magnetic field $B$ are $[MT^{-2}A^{-1}]$.
The dimensions of restoring torque constant $k$ are the same as torque,which is $[ML^2T^{-2}]$.
Substituting these dimensions into the formula: $[I_s] = \frac{[L^2] \cdot [MT^{-2}A^{-1}]}{[ML^2T^{-2}]}$.
Simplifying the expression: $[I_s] = \frac{L^2 MT^{-2}A^{-1}}{ML^2T^{-2}} = [A^{-1}]$.
Thus,the dimensional formula of current sensitivity is $[A^{-1}]$.

(C) Current sensitivity is defined as $I_s = \frac{NAB}{k}$,where $N$ is the number of turns,$A$ is the area,$B$ is the magnetic field,and $k$ is the torsional constant.
If $N$ is doubled,$I_s$ becomes $2 \times I_s$,meaning it is doubled.
Voltage sensitivity is defined as $V_s = \frac{I_s}{R} = \frac{NAB}{kR}$,where $R$ is the resistance of the coil.
Since the resistance $R$ is directly proportional to the length of the wire,and the length is proportional to the number of turns $(R \propto N)$,doubling $N$ also doubles $R$.
Substituting these into the formula: $V_s = \frac{(2N)AB}{k(2R)} = \frac{NAB}{kR}$.
Thus,the voltage sensitivity remains unchanged.

(A) To increase the range of a voltmeter from $V$ to $V'$,a high resistance $R$ must be connected in series with the voltmeter.
The formula for the required series resistance is $R = R_v (\frac{V'}{V} - 1)$,where $R_v$ is the internal resistance of the voltmeter.
Given: $R_v = x\text{ }\Omega$,$V = 20\text{ V}$,and $V' = 30\text{ V}$.
Substituting these values into the formula:
$R = x (\frac{30}{20} - 1)$
$R = x (1.5 - 1)$
$R = 0.5x = \frac{x}{2}\text{ }\Omega$.
Therefore,a resistor of $\frac{x}{2}\text{ }\Omega$ must be connected in series with the voltmeter.

(C) Let $G$ be the galvanometer resistance and $I_g$ be the current for full-scale deflection.
Case $1$: Shunt $S = 2\text{ }\Omega$,total current $I = 500\text{ mA} = 0.5\text{ A}$.
Using the shunt formula: $I_g G = (I - I_g)S$
$I_g G = (0.5 - I_g) \times 2$
$I_g(G + 2) = 1 \Rightarrow I_g = \frac{1}{G + 2}$ ... (Equation $1$)
Case $2$: Series resistance $R_s = 470\text{ }\Omega$,voltage $V = 10\text{ V}$.
Using Ohm's law: $I_g = \frac{V}{G + R_s}$
$I_g = \frac{10}{G + 470}$ ... (Equation $2$)
Equating Equation $1$ and Equation $2$:
$\frac{1}{G + 2} = \frac{10}{G + 470}$
$G + 470 = 10(G + 2)$
$G + 470 = 10G + 20$
$9G = 450$
$G = 50\text{ }\Omega$.

(D) The shunt resistance $S$ required to convert a galvanometer into an ammeter is given by the formula $S = \frac{I_g G}{I - I_g}$.
Here, $I_g$ is the current for full scale deflection, $G$ is the galvanometer resistance, and $I$ is the maximum current to be measured by the ammeter.
Given values are $I_g = 1 \text{ mA} = 10^{-3} \text{ A}$, $G = 100 \Omega$, and $I = 10 \text{ A}$.
Substituting these values into the formula:
$S = \frac{10^{-3} \times 100}{10 - 10^{-3}}$
$S = \frac{0.1}{9.999}$
$S \approx \frac{0.1}{10} = 0.01 \Omega$.
Thus, the required shunt resistance is approximately $0.01 \Omega$.