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

(B) In a suspended-type moving coil galvanometer,the coil is suspended by a thin fiber.
Quartz is preferred for this suspension because it exhibits negligible elastic after-effects.
This means that when the coil returns to its original position after a deflection,the fiber does not retain any residual deformation,ensuring high accuracy and sensitivity in measurements.
Therefore,the correct option is $B$.

(C) To convert a galvanometer into an ammeter,a low resistance,known as a shunt,must be connected in parallel with the galvanometer. This arrangement reduces the overall resistance of the device and allows the majority of the current to pass through the shunt,protecting the galvanometer coil from damage.

(D) To measure the potential difference across a component accurately,the voltmeter is connected in parallel with it.
If the voltmeter has a finite resistance,it draws some current from the circuit,which changes the potential difference being measured.
To minimize this error,the voltmeter must have a very high resistance compared to the resistance across which the potential difference is being measured.
An ideal voltmeter has an infinite resistance,so that it draws no current from the circuit.
Therefore,for a resistance of $1000 \ \Omega$,the voltmeter resistance should be much greater than $1000 \ \Omega$ (i.e.,$>> 1000 \ \Omega$).

(C) Given: Galvanometer resistance $G = 100\,\Omega$,full-scale deflection current $i_g = 10\,mA = 0.01\,A$,and desired range $i = 10\,A$.
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: $S = \frac{0.01 \times 100}{10 - 0.01} = \frac{1}{9.99} \approx 0.1\,\Omega$ (approximately).
Given the options,$0.1\,\Omega$ is the correct choice.

(C) An ammeter is an instrument used to measure the electric current flowing through a circuit.
It is always connected in series with the component through which the current is to be measured.
Therefore,the correct option is $C$.

(C) To convert a galvanometer (or ammeter) into a higher range ammeter,a shunt resistance $S$ is connected in parallel.
The formula for shunt resistance is given by $S = \frac{I_g G}{I - I_g}$,where:
$I_g = 1\, A$ (current capacity of the original ammeter)
$G = 0.018\,\Omega$ (resistance of the original ammeter)
$I = 10\, A$ (new range of the ammeter)
Substituting the values:
$S = \frac{1 \times 0.018}{10 - 1}$
$S = \frac{0.018}{9}$
$S = 0.002\,\Omega$
Therefore,the required shunt resistance is $0.002\,\Omega$.

(C) Let the main current be $I$ and the galvanometer resistance be $G = 99\,\Omega$.
Given that the current through the galvanometer is $I_g = 10\,\% \text{ of } I = 0.1I$.
The current through the shunt resistance $S$ is $I_s = I - I_g = I - 0.1I = 0.9I$.
Since the galvanometer and shunt are in parallel,the potential difference across them is equal:
$I_g G = I_s S$
$0.1I \times 99 = 0.9I \times S$
$S = \frac{0.1 \times 99}{0.9} = \frac{9.9}{0.9} = 11\,\Omega$.
Therefore,the required shunt resistance is $11\,\Omega$.

(D) To convert a galvanometer (or ammeter) into a voltmeter,a high resistance $R$ must be connected in series with it.
Given:
Resistance of the device $G = 5\,\Omega$
Full-scale current $I_g = 5\,mA = 5 \times 10^{-3}\,A$
Required voltage range $V = 100\,V$
The formula for series resistance is $V = I_g(R + G)$.
Rearranging for $R$: $R = \frac{V}{I_g} - G$.
Substituting the values:
$R = \frac{100}{5 \times 10^{-3}} - 5$
$R = 20,000 - 5$
$R = 19,995\,\Omega$.

(D) To convert a galvanometer into a voltmeter,a high resistance $R$ is connected in series with it.
The formula for the series resistance is $R = \frac{V}{I_g} - G$.
Given:
Full scale current $I_g = 100\,mA = 100 \times 10^{-3}\,A = 0.1\,A$.
Galvanometer resistance $G = 2\,\Omega$.
Voltage to be measured $V = 5\,V$.
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.

(C) The deflection of a galvanometer is directly proportional to the current passing through it,i.e.,$\theta \propto I$.
Initially,the current $I$ corresponds to $50$ divisions. When a shunt resistor $S = 12\,\Omega$ is connected in parallel,the current $I_g$ through the galvanometer corresponds to $10$ divisions.
Using the current divider rule for a galvanometer with resistance $G$ and shunt $S$:
$I_g = I \left( \frac{S}{S + G} \right)$
Given $\theta_1 = 50$ and $\theta_2 = 10$,we have $\frac{I_g}{I} = \frac{10}{50} = \frac{1}{5}$.
Substituting the values: $\frac{1}{5} = \frac{12}{12 + G}$.
$12 + G = 60$.
$G = 60 - 12 = 48\,\Omega$.
Therefore,the resistance of the galvanometer is $48\,\Omega$.

(A) To convert a galvanometer into an ammeter,a shunt resistance $S$ is connected in parallel with the galvanometer.
Given:
Galvanometer resistance $G = 10 \,\Omega$
Full scale deflection current $I_g = 0.01 \, A$
Range of ammeter $I = 10 \, A$
The formula for shunt resistance is $S = \frac{I_g G}{I - I_g}$.
Substituting the values:
$S = \frac{0.01 \times 10}{10 - 0.01}$
$S = \frac{0.1}{9.99}$
$S = \frac{10}{999} \,\Omega$.

(B) Let the main current be $I$ and the galvanometer resistance be $G = 90\, \Omega$.
Given that the current through the galvanometer $I_g = 10\% \text{ of } I = \frac{I}{10}$.
To allow only a fraction of the current to pass through the galvanometer,a shunt resistor $S$ must be connected in parallel with it.
The formula for the shunt resistance is $S = \frac{I_g \cdot G}{I - I_g}$.
Substituting the values: $S = \frac{(\frac{I}{10}) \cdot 90}{I - \frac{I}{10}} = \frac{9I}{0.9I} = \frac{90}{9} = 10\, \Omega$.
Thus,a resistor of $10\, \Omega$ should be connected in parallel.

(B) Let the resistance $R$ be connected in series with the voltmeter as shown in the diagram.
The initial range of the voltmeter is $V = i_g G$, where $i_g$ is the full-scale deflection current.
To increase the range to $V' = nV$, we connect a resistance $R$ in series.
The new total voltage is $V' = i_g(G + R)$.
Substituting $V' = nV$ and $i_g = \frac{V}{G}$:
$nV = \frac{V}{G}(G + R)$
$n = \frac{G + R}{G}$
$n = 1 + \frac{R}{G}$
$\frac{R}{G} = n - 1$
$R = (n - 1)G$

(C) An ammeter is a device used to measure the current flowing through a circuit. To measure the total current,it must be connected in series with the circuit component. If it were connected in parallel,it would create a short circuit due to its very low resistance,potentially damaging the device. Therefore,statement $C$ is incorrect because an ammeter is always connected in series,not in parallel.

(C) The voltmeter is connected in parallel to the combination of the ammeter and the resistor $R$.
Let the resistance of the ammeter be $r_A$.
Since the ammeter and resistor $R$ are in series,the total resistance of this branch is $(R + r_A)$.
The voltage across this branch is given by the voltmeter reading,$V = 20\, V$.
The current flowing through this branch is given by the ammeter reading,$I = 4\, A$.
According to Ohm's law,$V = I(R + r_A)$.
Substituting the given values,$20 = 4(R + r_A)$.
Dividing by $4$,we get $R + r_A = 5\,\Omega$.
Since the ammeter is not ideal,it must have some finite positive resistance,$r_A > 0$.
Therefore,$R = 5 - r_A$.
Since $r_A > 0$,it follows that $R < 5\,\Omega$.

(B) To convert a galvanometer into an ammeter,a shunt resistance $S$ must be connected in parallel with the galvanometer.
Given: Galvanometer resistance $G = 50\,\Omega$,full-scale current $i_g = 10\,mA = 10 \times 10^{-3}\,A$,and desired range $i = 1\,A$.
The formula for shunt resistance is $S = \frac{i_g \times G}{i - i_g}$.
Substituting the values: $S = \frac{10 \times 10^{-3} \times 50}{1 - 10 \times 10^{-3}} = \frac{0.5}{1 - 0.01} = \frac{0.5}{0.99} = \frac{50}{99}\,\Omega$.
Therefore,a resistance of $50/99\,\Omega$ must be connected in parallel.

(B) Let the initial current be $i$. The current through the coil $i_g$ is to be reduced by $90 \, \%$, so $i_g = i - 0.9i = 0.1i = \frac{i}{10}$.
Here, the ratio $n = \frac{i}{i_g} = 10$.
The resistance of the coil is $G = 900 \, \Omega$.
The formula for the shunt resistance $S$ required to reduce the current is $S = \frac{G}{n - 1}$.
Substituting the values, $S = \frac{900}{10 - 1} = \frac{900}{9} = 100 \, \Omega$.

(D) To convert a galvanometer into a voltmeter,a high resistance $R$ is connected in series with the galvanometer.
The formula for the series resistance is given by $R = \frac{V}{I_g} - G$,where:
$V = 100\,V$ (desired range of the voltmeter)
$I_g = 10\,mA = 10 \times 10^{-3}\,A = 0.01\,A$ (current for full-scale deflection)
$G = 25\,\Omega$ (resistance of the galvanometer)
Substituting the values:
$R = \frac{100}{0.01} - 25$
$R = 10000 - 25$
$R = 9975\,\Omega$
Therefore,the correct option is $D$.

(A) Given:
Galvanometer resistance,$G = 25\,\Omega$
Full-scale deflection current,$i_g = 50\,\mu A = 50 \times 10^{-6}\,A = 5 \times 10^{-5}\,A$
Required range of ammeter,$i = 5\,A$
The formula for shunt resistance $S$ is given by:
$S = \frac{G \cdot i_g}{i - i_g}$
Since $i_g$ is very small compared to $i$,we can approximate $i - i_g \approx i$ or use the exact formula:
$S = \frac{25 \times 5 \times 10^{-5}}{5 - 5 \times 10^{-5}}$
$S = \frac{125 \times 10^{-5}}{4.99995} \approx \frac{125 \times 10^{-5}}{5} = 25 \times 10^{-5} = 2.5 \times 10^{-4}\,\Omega$

(C) The error in measurement is given by the difference between the actual $e.m.f.$ and the measured terminal voltage.
Actual $e.m.f.$ $(E)$ = $2 \, V$.
The circuit consists of the cell (with internal resistance $r = 2 \, \Omega$) and the voltmeter (with resistance $R = 998 \, \Omega$) in series.
The current $(i)$ in the circuit is given by $i = \frac{E}{R + r} = \frac{2}{998 + 2} = \frac{2}{1000} = \frac{1}{500} \, A$.
The measured voltage $(V)$ across the voltmeter is the terminal voltage of the cell, given by $V = E - ir$.
Substituting the values: $V = 2 - (\frac{1}{500} \times 2) = 2 - \frac{2}{500} = 2 - 0.004 = 1.996 \, V$.
The error in measurement = $E - V = 2 - 1.996 = 0.004 \, V = 4 \times 10^{-3} \, V$.

(A) An ammeter is used to measure the electric current flowing through a circuit.
To measure current accurately without significantly altering the circuit's behavior,an ammeter must have a very low resistance.
Furthermore,to measure the current passing through a component,the ammeter must be connected in series with that component.
Therefore,the correct statement is that an ammeter has low resistance and is connected in series.

(B) Let the electromotive force $(EMF)$ of the battery be $E$.
When the ammeter is connected in series,the total resistance of the circuit is $R_{total} = R_1 + R_2 + R_A = 700\,\Omega + 410\,\Omega + 90\,\Omega = 1200\,\Omega$.
The current measured by the ammeter is $I_{measured} = E / R_{total} = 1.85\,A$.
Therefore,$E = 1.85\,A \times 1200\,\Omega = 2220\,V$.
When the ammeter is removed,the actual resistance of the circuit is $R_{actual} = R_1 + R_2 = 700\,\Omega + 410\,\Omega = 1110\,\Omega$.
The actual current is $I_{actual} = E / R_{actual} = 2220\,V / 1110\,\Omega = 2.0\,A$.
Since $2.0\,A > 1.85\,A$,the actual current is greater than the measured current.

(B) To convert a galvanometer into an ammeter,a shunt resistance $S$ must be connected in parallel with the galvanometer.
Given:
Galvanometer resistance $G = 50\,\Omega$
Full-scale deflection current $i_g = 100\,\mu A = 100 \times 10^{-6}\,A = 10^{-4}\,A$
Desired range of ammeter $i = 10\,A$
The formula for shunt resistance is $S = \frac{G \cdot i_g}{i - i_g}$.
Substituting the values:
$S = \frac{50 \times 10^{-4}}{10 - 10^{-4}}$
Since $10^{-4}$ is very small compared to $10$,we can approximate the denominator as $10$.
$S \approx \frac{50 \times 10^{-4}}{10} = 5 \times 10^{-4}\,\Omega$.
Thus,a resistance of $5 \times 10^{-4}\,\Omega$ must be connected in parallel.

(B) Let the main current be $I$ and the current through the galvanometer be $I_g$.
Given that $I_g = 2\%$ of $I$,so $I_g = \frac{2}{100} I = \frac{I}{50}$.
Let $S$ be the shunt resistance connected in parallel to the galvanometer of resistance $G$.
The formula for the current through the galvanometer is $I_g = I \left( \frac{S}{S + G} \right)$.
Substituting the values: $\frac{I}{50} = I \left( \frac{S}{S + G} \right)$.
$\frac{1}{50} = \frac{S}{S + G} \Rightarrow S + G = 50S$.
$G = 49S \Rightarrow S = \frac{G}{49}$.

(D) The resistance of a voltmeter is kept very high so that it draws negligible current from the circuit when connected in parallel across a component.
If the voltmeter had low resistance,it would draw a significant amount of current,which would alter the potential difference across the component being measured due to the internal resistance of the source or other circuit elements.
Therefore,a high resistance ensures that the measurement does not appreciably change the potential difference across the component.

(A) To convert a galvanometer into a voltmeter,a high resistance $R$ must be connected in series with the galvanometer.
Given:
Galvanometer resistance $G = 7\,\Omega$
Full-scale deflection current $i_g = 1.0\,A$
Desired voltage range $V = 10\,V$
The formula for the series resistance $R$ is:
$R = \frac{V}{i_g} - G$
Substituting the values:
$R = \frac{10}{1.0} - 7 = 10 - 7 = 3\,\Omega$
Therefore,a resistance of $3\,\Omega$ must be connected in series with the galvanometer.

(B) Let $G = 9 \, \Omega$ be the resistance of the galvanometer and $S = 2 \, \Omega$ be the resistance of the shunt.
Let $I = 1 \, A$ be the total current.
Let $I_s$ be the current passing through the shunt and $I_g$ be the current passing through the galvanometer.
According to the principle of parallel circuits, the voltage across the galvanometer and the shunt is the same: $V = I_g G = I_s S$.
Since $I = I_g + I_s$, we have $I_g = I - I_s$.
Substituting this into the voltage equation: $(I - I_s) G = I_s S$.
Rearranging to solve for $I_s$: $I G - I_s G = I_s S \implies I G = I_s (S + G)$.
Therefore, $I_s = I \times \frac{G}{S + G}$.
Substituting the given values: $I_s = 1 \times \frac{9}{2 + 9} = \frac{9}{11} \approx 0.818 \, A$.
Rounding to the nearest provided option, the current through the shunt is approximately $0.8 \, A$.

(B) Given: Galvanometer resistance $G = 180\,\Omega$,full-scale deflection current $i_g = 2\,mA$,and the desired range $i = 20\,mA$.
To convert a galvanometer into an ammeter,a shunt resistance $S$ is connected in parallel.
The formula for the shunt resistance is given by $S = \frac{i_g G}{i - i_g}$.
Substituting the values: $S = \frac{2\,mA \times 180\,\Omega}{20\,mA - 2\,mA}$.
$S = \frac{2 \times 180}{18} = \frac{360}{18} = 20\,\Omega$.
Therefore,the required shunt resistance is $20\,\Omega$.

(C) The resistance of the galvanometer is $G = 120 \, \Omega$ and the full-scale deflection current is $i_g = 0.05 \, A$.
The maximum current to be measured is $i = 10 \, A$.
The shunt resistance $S$ is connected in parallel to the galvanometer.
The resistance of the resulting ammeter $R_A$ is given by the parallel combination of $G$ and $S$: $R_A = \frac{G \cdot S}{G + S}$.
From the principle of shunting,the voltage across the galvanometer and shunt is equal: $i_g \cdot G = (i - i_g) \cdot S$.
Thus,$S = \frac{i_g \cdot G}{i - i_g} = \frac{0.05 \times 120}{10 - 0.05} = \frac{6}{9.95} \approx 0.603 \, \Omega$.
Alternatively,the equivalent resistance $R_A$ is simply the voltage across the parallel combination divided by the total current: $R_A = \frac{V}{i} = \frac{i_g \cdot G}{i} = \frac{0.05 \times 120}{10} = \frac{6}{10} = 0.6 \, \Omega$.

(D) Given: Resistance of the device $G = 1000\,\Omega$,full-scale current $I_g = 100\,mA = 0.1\,A$,and the desired range $I = 1\,A$.
To convert a galvanometer (or voltmeter) 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: $S = \frac{0.1 \times 1000}{1 - 0.1}$.
$S = \frac{100}{0.9} = \frac{1000}{9} \approx 111.11\,\Omega$.
Rounding to the nearest integer,we get $111\,\Omega$.

(A) Let $G$ be the resistance of the galvanometer and $S$ be the shunt resistance.
Given: $G = 25\,\Omega$ and $S = 2.5\,\Omega$.
Let $I$ be the total current and $I_g$ be the current flowing through the galvanometer.
According to the current divider rule,the current through the galvanometer is given by:
$I_g = I \times \frac{S}{G + S}$
Therefore,the ratio of the current through the galvanometer to the total current is:
$\frac{I_g}{I} = \frac{S}{G + S}$
Substituting the given values:
$\frac{I_g}{I} = \frac{2.5}{25 + 2.5} = \frac{2.5}{27.5}$
$\frac{I_g}{I} = \frac{25}{275} = \frac{1}{11}$

(D) Given: Resistance of the voltmeter $G = 2000 \, \Omega$, initial range $V_g = 2 \, V$, and desired range $V = 10 \, V$.
To convert a voltmeter to a higher range, a resistance $R$ is connected in series with the voltmeter.
The formula for the series resistance is $R = G \left( \frac{V}{V_g} - 1 \right)$.
Substituting the values: $R = 2000 \left( \frac{10}{2} - 1 \right)$.
$R = 2000 (5 - 1) = 2000 \times 4 = 8000 \, \Omega$.
Therefore, the required resistance is $8000 \, \Omega$.

(C) Let the full-scale deflection current of the galvanometer be $i_g$ and its resistance be $R$.
To convert the galvanometer into an ammeter of range $n$ times the original range,the required shunt resistance $S$ is given by the formula:
$S = \frac{R}{n-1}$
Here,the range is $4$ times the original range,so $n = 4$.
Substituting the value of $n$ into the formula:
$S = \frac{R}{4-1} = \frac{R}{3}$.
Therefore,the required shunt resistance is $\frac{R}{3}$.

(A) An ammeter is designed to have a very low resistance so that it can measure current without significantly altering the circuit.
When an ammeter is connected in parallel to a circuit element,the effective resistance of that branch becomes extremely low.
According to Ohm's law,$I = V/R$,a very low resistance leads to a very high current flow through the ammeter.
This excessive current exceeds the rated capacity of the ammeter,which is likely to cause damage to the device.

(B) To convert a galvanometer into a voltmeter,a high resistance $R$ must be connected in series with the galvanometer.
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,and $G$ is the galvanometer resistance.
Given: $V = 18 \, V$,$I_g = 3 \, mA = 3 \times 10^{-3} \, A$,and $G = 12 \, \Omega$.
Substituting the values: $R = \frac{18}{3 \times 10^{-3}} - 12$.
$R = 6000 - 12 = 5988 \, \Omega$.
Thus,a resistance of $5988 \, \Omega$ must be connected in series.

(D) An ammeter is a device connected in series to a circuit to measure the current flowing through it.
If an ammeter has significant resistance,it increases the total equivalent resistance of the circuit,which in turn reduces the current being measured,leading to an error.
In practice,no conductor has zero resistance,but practical ammeters are designed to have very low resistance to minimize this error.
For an ideal or perfect ammeter,the resistance must be $0 \ \Omega$ so that the measuring instrument does not alter the circuit's characteristics or the current it is intended to measure.

(C) To convert a galvanometer into a voltmeter,a high resistance $R$ must be connected in series with the galvanometer.
Given: Galvanometer resistance $G = 25 \,\Omega$,full-scale deflection current $i_g = 6 \,mA = 6 \times 10^{-3} \,A$,and maximum voltage to be measured $V = 6 \,V$.
The formula for the series resistance $R$ is given by $R = \frac{V}{i_g} - G$.
Substituting the values: $R = \frac{6}{6 \times 10^{-3}} - 25$.
$R = 1000 - 25 = 975 \,\Omega$.
Thus,a resistance of $975 \,\Omega$ must be connected in series.

(D) The formula for the shunt resistance $S$ required to convert a galvanometer of resistance $G$ into an ammeter of range $I$ is given by $S = \frac{I_g G}{I - I_g}$, where $I_g$ is the maximum current the galvanometer can measure.
Given values are $G = 25 \, \Omega$, $I_g = 0.01 \, A$, and $I = 10 \, A$.
Substituting these values into the formula:
$S = \frac{0.01 \times 25}{10 - 0.01}$
$S = \frac{0.25}{9.99}$
$S = \frac{25}{999} \, \Omega$.
Thus, the required shunt resistance is $25/999 \, \Omega$.

(A) The full-scale deflection current $i_g$ of the galvanometer is given by:
$i_g = \text{number of divisions} \times \text{sensitivity}$
$i_g = 30 \times 16 \times 10^{-6} \, A = 480 \times 10^{-6} \, A = 4.8 \times 10^{-4} \, A$
To convert a galvanometer into a voltmeter,a high resistance $R$ is connected in series with it.
The formula for the total resistance is:
$V = i_g(R + G)$
Assuming the internal resistance $G$ of the galvanometer is negligible compared to the high series resistance $R$ (or considering $R+G \approx R$):
$R \approx \frac{V}{i_g} = \frac{3}{4.8 \times 10^{-4}} = \frac{30000}{4.8} = 6250 \, \Omega = 6.25 \, k\Omega$
Thus,a resistance of nearly $6 \, k\Omega$ connected in series is required.

(D) The resistance of voltmeter $V_1$ is $R_1 = 80 \, V \times 200 \, \Omega/V = 16000 \, \Omega = 16 \, k\Omega$.
Since the voltmeters are connected in series, the current $I$ flowing through both is the same.
$I = \frac{V_1}{R_1} = \frac{80 \, V}{16 \times 10^3 \, \Omega} = 5 \times 10^{-3} \, A$.
The potential difference across $V_2$ is $V_2 = I \times R_2 = (5 \times 10^{-3} \, A) \times (32 \times 10^3 \, \Omega) = 160 \, V$.
The total line voltage $V$ is the sum of the potential differences across the series components:
$V = V_1 + V_2 = 80 \, V + 160 \, V = 240 \, V$.

(C) To convert an ammeter (or galvanometer) to measure a higher current range,a shunt resistance $S$ must be connected in parallel with it.
Given:
Full-scale current of ammeter,$I_g = 2 \, A$
Resistance of ammeter,$G = 12 \, \Omega$
Maximum current to be measured,$I = 5 \, A$
The formula for shunt resistance is given by:
$S = \frac{I_g \cdot G}{I - I_g}$
Substituting the values:
$S = \frac{2 \cdot 12}{5 - 2}$
$S = \frac{24}{3} = 8 \, \Omega$
Therefore,a resistance of $8 \, \Omega$ must be connected in parallel.

(D) Let the total current in the circuit be $I$ and the current through the galvanometer be $I_g$.
Given that $I_g = 5\% \text{ of } I = \frac{5}{100} I = \frac{1}{20} I$.
The formula for the current through a galvanometer in parallel with a shunt resistance $S$ is given by $I_g = I \left( \frac{S}{G + S} \right)$.
Substituting the values: $\frac{1}{20} I = I \left( \frac{S}{G + S} \right)$.
$\frac{1}{20} = \frac{S}{G + S}$.
$G + S = 20S$.
$G = 19S$.
Therefore,$S = \frac{G}{19}$.

(A) The resistance of the voltmeter is given as $G = 50 \times 10^3 \, \Omega$.
To increase the range of a voltmeter by a factor of $n$,we connect a resistance $R$ in series with the voltmeter.
The formula for the required series resistance is $R = G(n - 1)$.
Here,the range is to be increased $3$ times,so $n = 3$.
Substituting the values: $R = (50 \times 10^3 \, \Omega) \times (3 - 1)$.
$R = 50 \times 10^3 \times 2 = 100 \times 10^3 \, \Omega = 10^5 \, \Omega$.

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

(D) For a galvanometer with resistance $G$ to be converted into a voltmeter,a resistance $R$ is connected in series. The full-scale deflection current $i_g$ remains constant.
For the first case: $i_g = \frac{V}{R + G}$
For the second case: $i_g = \frac{V'}{2R + G}$
Equating the two expressions for $i_g$: $\frac{V}{R + G} = \frac{V'}{2R + G}$
Rearranging for $V'$: $V' = V \times \frac{2R + G}{R + G}$
$V' = V \times \frac{2(R + G) - G}{R + G} = V \times \left( 2 - \frac{G}{R + G} \right)$
$V' = 2V - \frac{VG}{R + G}$
Since $\frac{VG}{R + G} > 0$,it follows that $V' < 2V$.

(C) Let $G$ be the resistance of the galvanometer and $S$ be the shunt resistance.
Given: $G = 36 \,\Omega$ and $S = 4 \,\Omega$.
The current $i_g$ passing through the galvanometer is related to the total current $i$ by the formula:
$i_g = i \left( \frac{S}{G + S} \right)$
Substituting the given values:
$\frac{i_g}{i} = \frac{4}{36 + 4} = \frac{4}{40} = \frac{1}{10}$
To find the percentage,multiply by $100$:
$\text{Percentage} = \frac{1}{10} \times 100 = 10\%$.
Thus,$10\%$ of the total current passes through the galvanometer.

(C) The resistance of the galvanometer is $G = 20 \,\Omega$.
The full-scale deflection current is $I_g = 1 \, mA = 10^{-3} \, A$.
The desired range of the ammeter is $I = 1 \, A$.
To convert a galvanometer into an ammeter,a shunt resistance $S$ is connected in parallel with the galvanometer.
The formula for the shunt resistance is $S = \frac{I_g \cdot G}{I - I_g}$.
Substituting the values: $S = \frac{10^{-3} \times 20}{1 - 10^{-3}} = \frac{0.02}{0.999} \approx 0.02 \,\Omega$.

(A) voltmeter is used to measure the potential difference between two points in a circuit.
To ensure that the voltmeter does not draw significant current from the circuit and does not alter the potential difference being measured,it must have a very high resistance.
An ideal voltmeter has infinite resistance.
Therefore,among the given options,the voltmeter with the highest resistance,which is $10000\,\Omega$,is the best.

(C) An ammeter is designed to have a very low resistance so that it can measure current without significantly altering the circuit. $A$ voltmeter is designed to have a very high resistance to ensure it draws negligible current from the circuit.
If an ammeter is connected in parallel (as a voltmeter is),the low resistance of the ammeter would cause a very large current to flow through it,potentially damaging the device.
To convert an ammeter into a voltmeter,we must increase its total resistance to a very high value. This is achieved by connecting a high resistance in series with the ammeter.

(A) The millivoltmeter has a full-scale deflection voltage $V_g = 800\, mV = 0.8\, V$ and internal resistance $G = 40\,\Omega$.
To convert this into an ammeter of range $I = 100\, mA = 0.1\, A$,we connect a shunt resistance $S$ in parallel.
The current through the galvanometer is $I_g = \frac{V_g}{G} = \frac{0.8}{40} = 0.02\, A$.
The current through the shunt is $I_s = I - I_g = 0.1 - 0.02 = 0.08\, A$.
Since the shunt is in parallel with the galvanometer,the voltage across them is equal: $I_g \times G = I_s \times S$.
$0.02 \times 40 = 0.08 \times S$.
$0.8 = 0.08 \times S$.
$S = \frac{0.8}{0.08} = 10\,\Omega$.