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

(B) Given: Resistance of galvanometer $G = 8 \, \Omega$,Shunt resistance $S = 2 \, \Omega$,Total current $I = 1 \, A$.
The current through the shunt $I_s$ is given by the formula:
$I_s = I \times \left( \frac{G}{S + G} \right)$
Substituting the values:
$I_s = 1 \times \left( \frac{8}{2 + 8} \right)$
$I_s = \frac{8}{10} = 0.8 \, A$.
Therefore,the current passing through the shunt is $0.8 \, A$.

(B) The formula for the shunt resistance $S$ required to convert a galvanometer of resistance $G$ and full-scale deflection current $i_g$ into an ammeter of range $i$ is given by: $S = \frac{i_g}{i - i_g} G$,which implies $i_g G = (i - i_g) S$.
For the first case,with shunt $S_1 = 4r$ and range $i_1 = 0.03 \, A$:
$i_g G = (0.03 - i_g) 4r$ --- $(i)$
For the second case,with shunt $S_2 = r$ and range $i_2 = 0.06 \, A$:
$i_g G = (0.06 - i_g) r$ --- $(ii)$
Equating $(i)$ and $(ii)$:
$(0.03 - i_g) 4r = (0.06 - i_g) r$
$4(0.03 - i_g) = 0.06 - i_g$
$0.12 - 4i_g = 0.06 - i_g$
$3i_g = 0.06$
$i_g = 0.02 \, A$.

(D) The current sensitivity $(S_i)$ of a galvanometer is given by the formula: $S_i = \frac{\theta}{i} = \frac{NBA}{k}$,where $N$ is the number of turns,$B$ is the magnetic field,$A$ is the area,and $k$ is the restoring torque constant.
Given: $N = 100$,$A = 1 \, cm^2 = 10^{-4} \, m^2$,$B = 5 \, T$,and $k = 10^{-8} \, N-m/rad$.
Substituting the values:
$S_i = \frac{100 \times 5 \times 10^{-4}}{10^{-8}}$
$S_i = \frac{500 \times 10^{-4}}{10^{-8}} = 500 \times 10^4 = 5 \times 10^6 \, rad/A$.
Since $1 \, A = 10^6 \, \mu A$,the sensitivity is $5 \, rad/\mu A$.

(C) The circuit is shown in the figure.
First,calculate the effective resistance of the ammeter $(R_A)$. Since the coil resistance $(480\,\Omega)$ and the shunt resistance $(20\,\Omega)$ are in parallel,the equivalent resistance is:
$R_A = \frac{480 \times 20}{480 + 20} = \frac{9600}{500} = 19.2\,\Omega$
Since the ammeter is connected in series with the $40.8\,\Omega$ resistor,the total resistance $(R_{total})$ of the circuit is:
$R_{total} = 40.8\,\Omega + 19.2\,\Omega = 60\,\Omega$
Using Ohm's law $(I = V/R)$,the total current flowing through the circuit is:
$I = \frac{30\,V}{60\,\Omega} = 0.5\,A$
Thus,the reading in the ammeter will be $0.5\,A$.

(D) Current sensitivity $(I_s)$ is defined as the deflection per unit current: $I_s = \frac{\theta}{I}$.
Voltage sensitivity $(V_s)$ is defined as the deflection per unit voltage: $V_s = \frac{\theta}{V}$.
Since $V = IR$,where $R$ is the resistance of the galvanometer $(R_G)$,we have $V_s = \frac{\theta}{I R_G} = \frac{I_s}{R_G}$.
Therefore,the resistance of the galvanometer is $R_G = \frac{I_s}{V_s}$.
Given: $I_s = 5\,div/mA = 5\,div/(10^{-3}\,A) = 5000\,div/A$.
Given: $V_s = 20\,div/V$.
Substituting the values: $R_G = \frac{5000}{20} = 250\,\Omega$.

(A) Let the shunt resistance be $S$.
Given:
Total current to be measured,$I = 750\, A$
Full-scale deflection current of the ammeter,$I_g = 100\, A$
Resistance of the ammeter,$R_G = 13\, \Omega$
When a shunt $S$ is connected in parallel with the ammeter,the potential difference across the ammeter and the shunt must be equal:
$I_g R_G = (I - I_g) S$
Substituting the given values:
$100 \times 13 = (750 - 100) \times S$
$1300 = 650 \times S$
Solving for $S$:
$S = \frac{1300}{650} = 2\, \Omega$
Thus,the value of the shunt resistance is $2\, \Omega$.

(B) The initial total resistance is $R_{total,1} = R_G + R_1 = 50\, \Omega + 2950\, \Omega = 3000\, \Omega$.
The current for full-scale deflection ($30$ divisions) is $I_1 = \frac{V}{R_{total,1}} = \frac{3\, V}{3000\, \Omega} = 1 \times 10^{-3}\, A = 1\, mA$.
To reduce the deflection to $20$ divisions,the new current $I_2$ must be proportional to the divisions: $I_2 = I_1 \times \frac{20}{30} = 1\, mA \times \frac{2}{3} = \frac{2}{3}\, mA$.
Using Ohm's law for the new circuit,$V = I_2 \times R_{total,2}$,where $R_{total,2} = R_G + R_2$.
$3\, V = (\frac{2}{3} \times 10^{-3}\, A) \times R_{total,2}$.
$R_{total,2} = \frac{3 \times 3}{2} \times 10^3\, \Omega = 4500\, \Omega$.
Since $R_{total,2} = R_G + R_2$,the required series resistance is $R_2 = 4500\, \Omega - 50\, \Omega = 4450\, \Omega$.

(C) To convert a galvanometer into an ammeter,a shunt resistance $S$ must be connected in parallel with the galvanometer.
The formula for the shunt resistance is $i_g G = (I - i_g) S$,where $i_g$ is the full-scale deflection current,$G$ is the galvanometer resistance,and $I$ is the maximum current to be measured.
Given: $G = 60\,\Omega$,$i_g = 1.0\,A$,and $I = 5.0\,A$.
Substituting the values: $1.0 \times 60 = (5.0 - 1.0) \times S$.
$60 = 4 \times S$.
$S = \frac{60}{4} = 15\,\Omega$.
Therefore,a resistance of $15\,\Omega$ must be connected in parallel.

(A) Resistance of galvanometer,$G = 100 \, \Omega$.
Current for full-scale deflection,$I_g = 30 \, mA = 30 \times 10^{-3} \, A$.
Range of voltmeter,$V = 30 \, V$.
To convert the galvanometer into a voltmeter of a given range,a resistance $R$ is connected in series with it.
The total resistance of the circuit is $(G + R)$.
According to Ohm's law,$V = I_g(G + R)$.
Substituting the values: $30 = 30 \times 10^{-3} \times (100 + R)$.
$1000 = 100 + R$.
$R = 1000 - 100 = 900 \, \Omega$.

(D) Let the original resistance of the galvanometer be $G$. When a shunt resistance $S$ is connected in parallel with the galvanometer,the equivalent resistance of this parallel combination is $R_p = \frac{GS}{G+S}$.
To keep the main current in the circuit unchanged,the total resistance of the circuit must remain equal to the original resistance of the galvanometer,which is $G$.
Let $R$ be the resistance to be connected in series with the parallel combination of $G$ and $S$.
Therefore,the total resistance is $R + R_p = G$.
Substituting the value of $R_p$,we get $R + \frac{GS}{G+S} = G$.
Solving for $R$,we have $R = G - \frac{GS}{G+S}$.
Taking the common denominator,$R = \frac{G(G+S) - GS}{G+S} = \frac{G^2 + GS - GS}{G+S}$.
Thus,$R = \frac{G^2}{G+S}$.

(A) The shunt resistance $S$ required to convert a galvanometer (or millivoltmeter) into an ammeter is given by the formula: $S = \frac{V_g}{I - I_g}$.
Here,$V_g$ is the full-scale voltage of the millivoltmeter,$I$ is the desired range of the ammeter,and $I_g$ is the full-scale current of the millivoltmeter.
Given $V_g = 25 \, mV = 25 \times 10^{-3} \, V$ and $I = 25 \, A$.
Since $I \gg I_g$,we can approximate $I - I_g \approx I$.
Therefore,$S = \frac{V_g}{I} = \frac{25 \times 10^{-3} \, V}{25 \, A} = 10^{-3} \, \Omega = 0.001 \, \Omega$.

(C) Let the total current be $I$ and the resistance of the galvanometer be $G$.
The current through the galvanometer is $I_G = 0.2\% \text{ of } I = \frac{0.2}{100} I = \frac{1}{500} I$.
The current through the shunt resistance $S$ is $I_S = I - I_G = I - \frac{1}{500} I = \frac{499}{500} I$.
Since the galvanometer and shunt are in parallel, the potential difference across them is the same:
$I_G G = I_S S$
$\left( \frac{1}{500} I \right) G = \left( \frac{499}{500} I \right) S$
$S = \frac{G}{499}$.
The resistance of the ammeter $R_A$ is the equivalent resistance of the parallel combination of $G$ and $S$:
$\frac{1}{R_A} = \frac{1}{G} + \frac{1}{S} = \frac{1}{G} + \frac{499}{G} = \frac{500}{G}$.
Therefore, $R_A = \frac{G}{500}$.

(B) Let $G$ be the resistance of the galvanometer and $i_g$ be the maximum current that can pass through it without any shunt.
The formula for the range of an ammeter with a shunt $S$ is given by $i_g G = (i - i_g)S$,where $i$ is the total current.
For the first case,$S = 4r$ and $i = 0.03\,A$:
$i_g G = (0.03 - i_g)4r$ ..... $(i)$
For the second case,$S = r$ and $i = 0.06\,A$:
$i_g G = (0.06 - i_g)r$ ..... $(ii)$
Equating $(i)$ and $(ii)$:
$(0.03 - i_g)4r = (0.06 - i_g)r$
$0.12 - 4i_g = 0.06 - i_g$
$0.06 = 3i_g$
$i_g = 0.02\,A$.

(C) The current sensitivity of a moving coil galvanometer is given by the formula:
$S_i = \frac{NBA}{C}$
where $N$ is the number of turns,$B$ is the magnetic field,$A$ is the area,and $C$ is the torsional constant of the spring.
From the formula,we see that $S_i \propto N$ when $B, A,$ and $C$ are kept constant.
Let the initial sensitivity be $(S_i)_1$ and the final sensitivity be $(S_i)_2$.
Given that the sensitivity increases by $25\%$,we have:
$(S_i)_2 = (S_i)_1 + 0.25(S_i)_1 = 1.25(S_i)_1$
Using the proportionality $S_i \propto N$:
$\frac{(S_i)_1}{(S_i)_2} = \frac{N_1}{N_2}$
$\frac{(S_i)_1}{1.25(S_i)_1} = \frac{48}{N_2}$
$\frac{1}{1.25} = \frac{48}{N_2}$
$N_2 = 48 \times 1.25 = 60$
Therefore,the number of turns should become $60$.

(D) Let the total current entering the circuit be $I$. When the galvanometer is shunted by a resistor $S = 20\,\Omega$,the current through the galvanometer $I_g$ becomes half of the total current,i.e.,$I_g = I/2$.
Using the principle of current division in a parallel circuit,the current through the galvanometer is given by:
$I_g = I \left( \frac{S}{S + R_g} \right)$
Substituting the given values:
$\frac{I}{2} = I \left( \frac{20}{20 + R_g} \right)$
$\frac{1}{2} = \frac{20}{20 + R_g}$
$20 + R_g = 40$
$R_g = 40 - 20 = 20\,\Omega$
Thus,the galvanometer resistance is $20\,\Omega$.

(B) Let $G$ be the resistance of the galvanometer and $i$ be the main current.
When shunted with $S_1 = 4\,\Omega$,the current through the galvanometer is $i_{g1} = i/5$.
Using the shunt formula: $i_{g1} = i \cdot \frac{S_1}{G + S_1} \implies \frac{i}{5} = i \cdot \frac{4}{G + 4}$.
Solving for $G$: $G + 4 = 20 \implies G = 16\,\Omega$.
Now,the galvanometer is further shunted with $S_2 = 2\,\Omega$ in parallel with $S_1 = 4\,\Omega$.
The equivalent shunt resistance $S_{eq}$ is: $\frac{1}{S_{eq}} = \frac{1}{4} + \frac{1}{2} = \frac{1+2}{4} = \frac{3}{4} \implies S_{eq} = \frac{4}{3}\,\Omega$.
The new current through the galvanometer is $i_{g2} = i \cdot \frac{S_{eq}}{G + S_{eq}} = i \cdot \frac{4/3}{16 + 4/3} = i \cdot \frac{4/3}{52/3} = \frac{4}{52}i = \frac{1}{13}i$.
The ratio of the new deflection to the previous deflection is $\frac{i_{g2}}{i_{g1}} = \frac{i/13}{i/5} = \frac{5}{13}$.
Thus,the new deflection is $\frac{5}{13}$ of the deflection when shunted with $4\,\Omega$ only.

(B) Given: Galvanometer resistance $R_g = 20\,\Omega$,full-scale voltage $V_g = 0.2\,V$,and target current $I = 10\,A$.
First,calculate the full-scale deflection current $I_g$ of the galvanometer:
$I_g = \frac{V_g}{R_g} = \frac{0.2\,V}{20\,\Omega} = 0.01\,A$.
To convert a galvanometer into an ammeter of range $I$,a shunt resistance $S$ must be connected in parallel with the galvanometer coil.
The formula for shunt resistance is:
$S = \frac{I_g R_g}{I - I_g}$.
Substituting the values:
$S = \frac{0.01 \times 20}{10 - 0.01} = \frac{0.2}{9.99} \approx 0.02\,\Omega$.
Therefore,a $0.02\,\Omega$ resistor should be connected in parallel (across) the galvanometer.

(C) The milliammeter has a full-scale deflection current $i_{g} = 10\, mA = 0.01\, A$ and resistance $R_{g} = 9\, \Omega$.
When terminals $A$ and $B$ are used,the current $I$ enters at $A$. The current $i_{g}$ flows through the milliammeter,and the remaining current $(I - i_{g})$ flows through the shunt resistor of $0.1\, \Omega$ connected between $A$ and $B$.
Since the milliammeter and the shunt resistor are in parallel between $A$ and $B$,the potential difference across them is equal.
$V_{AB} = i_{g} \times R_{g} = (I - i_{g}) \times R_{shunt}$
$0.01\, A \times 9\, \Omega = (I - 0.01\, A) \times 0.1\, \Omega$
$0.09 = 0.1I - 0.001$
$0.1I = 0.091$
$I = 0.91\, A = 910\, mA$.
Wait,re-evaluating the circuit: The milliammeter is in series with the $0.9\, \Omega$ resistor,and this combination is in parallel with the $0.1\, \Omega$ resistor.
$V_{AB} = i_{g}(R_{g} + 0.9\, \Omega) = (I - i_{g}) \times 0.1\, \Omega$
$0.01(9 + 0.9) = (I - 0.01) \times 0.1$
$0.01(9.9) = 0.1I - 0.001$
$0.099 + 0.001 = 0.1I$
$0.1 = 0.1I$
$I = 1\, A$.

(C) The full-scale voltage $V$ of the voltmeter is given by $V = I_g(R_g + R_s)$,where $I_g = 10\, mA = 0.01\, A$,$R_g = 90\, \Omega$,and $R_s = 910\, \Omega$.
$V = 0.01\, A \times (90\, \Omega + 910\, \Omega) = 0.01 \times 1000 = 10\, V$.
Given the least count of the voltmeter is $0.1\, V/\text{division}$.
The total number of divisions $N$ is given by $N = \frac{V}{\text{least count}}$.
$N = \frac{10\, V}{0.1\, V/\text{division}} = 100\, \text{divisions}$.

(D) To convert a galvanometer into an ammeter,a very small resistance (shunt) is connected in parallel. Thus,the resistance of an ammeter is very low.
To convert a galvanometer into a voltmeter,a very large resistance is connected in series. Thus,the resistance of a voltmeter is very high.
For a voltmeter,the resistance $R$ is given by $R = (V / I_g) - G$,where $V$ is the range,$I_g$ is the full-scale deflection current $(1 \, mA)$,and $G$ is the galvanometer resistance.
Since $R$ is directly proportional to the voltage range $V$,the voltmeter with the highest range will have the largest resistance.
Comparing the two voltmeters ($5 \, V$ and $10 \, V$),the voltmeter with a range of $10 \, V$ will have the largest resistance.

(D) The smaller the resistance of an ammeter,the more accurate its reading will be. An ammeter is considered ideal if its resistance is zero,meaning it measures the actual current flowing through the circuit without affecting it.
Similarly,the greater the resistance of a voltmeter,the more accurate its reading will be. $A$ voltmeter is considered ideal if its resistance is infinite,meaning it draws no current from the circuit element for its operation.
Therefore,an ammeter should have small resistance and a voltmeter should have large resistance.
Thus,both $(A)$ and $(C)$ are correct.

(D) Given: Galvanometer resistance $G = 100\,\Omega$,full-scale current $I_g = 50\,\mu A = 50 \times 10^{-6}\,A$.
For Voltmeter conversion: $A$ high resistance $R$ is connected in series. The formula is $R = \frac{V}{I_g} - G$.
For option $(B)$: $V = 10\,V$. $R = \frac{10}{50 \times 10^{-6}} - 100 = 200,000 - 100 \approx 200\,k\Omega$. Thus,$(B)$ is correct.
For Ammeter conversion: $A$ shunt resistance $S$ is connected in parallel. The formula is $S = \frac{I_g \cdot G}{I - I_g}$.
For option $(C)$: $I = 5\,mA = 5 \times 10^{-3}\,A$. $S = \frac{50 \times 10^{-6} \times 100}{5 \times 10^{-3} - 50 \times 10^{-6}} = \frac{5 \times 10^{-3}}{4.95 \times 10^{-3}} \approx 1.01\,\Omega \approx 1\,\Omega$. Thus,$(C)$ is correct.
Since both $(B)$ and $(C)$ are correct,the answer is $(D)$.

(D) The resistance of an ammeter is given by $R_A = \frac{G \cdot S}{G + S}$,where $G$ is the galvanometer resistance and $S$ is the shunt resistance.
To increase the range $(I)$ of an ammeter,the shunt resistance $S$ must be decreased,as $S = \frac{I_g G}{I - I_g}$.
As the range $I$ increases,the required shunt resistance $S$ decreases,which in turn decreases the total resistance of the ammeter $R_A$. Therefore,Statement-$I$ is false.
Statement-$II$ is true because connecting an additional shunt in parallel effectively reduces the total resistance and allows a larger current to bypass the galvanometer,thereby increasing the range.

(C) To convert a galvanometer into an ammeter,a shunt resistance $S$ is connected in parallel with the galvanometer.
Given:
Galvanometer resistance $G = 100 \ \Omega$
Full-scale deflection current $I_g = 1 \ mA = 10^{-3} \ A$
Total current to be measured $I = 10 \ A$
The formula for shunt resistance is $S = \frac{I_g G}{I - I_g}$.
Substituting the values:
$S = \frac{10^{-3} \times 100}{10 - 10^{-3}}$
$S = \frac{0.1}{9.999}$
$S \approx 0.01 \ \Omega$.

(A) Given: Current through the galvanometer,$i_{g} = 5 \times 10^{-3} \ A$.
Galvanometer resistance,$G = 15 \ \Omega$.
To convert a galvanometer into a voltmeter of range $V$,a high resistance $R$ must be connected in series with it.
The formula is $V = i_{g}(R + G)$.
Substituting the given values: $10 = 5 \times 10^{-3} \times (R + 15)$.
$R + 15 = \frac{10}{5 \times 10^{-3}} = 2000$.
$R = 2000 - 15 = 1985 \ \Omega$.
$R = 1.985 \times 10^{3} \ \Omega$.

(C) The resistance of the galvanometer $G$ is given by $G = \frac{V_g}{I_g} = \frac{0.15 \ V}{50 \times 10^{-3} \ A} = 3 \ \Omega$.
To convert the galvanometer into an ammeter of range $I = 100 \ A$,we connect a shunt resistance $S$ in parallel.
The formula for shunt resistance is $S = \frac{I_g G}{I - I_g}$.
Substituting the values: $S = \frac{50 \times 10^{-3} \times 3}{100 - 50 \times 10^{-3}}$.
$S = \frac{0.15}{100 - 0.05} = \frac{0.15}{99.95} \approx 0.0015 \ \Omega$.

(A) To construct an ammeter from a galvanometer,a shunt resistance $R_S$ is connected in parallel with the galvanometer.
The full-scale deflection current of the galvanometer $I_g$ is given by $I_g = V / r_g = 0.50 \ V / 25 \ \Omega = 0.02 \ A$.
The total current $I$ to be measured is $2.0 \ A$. The current through the shunt resistance $I_S$ is $I_S = I - I_g = 2.0 \ A - 0.02 \ A = 1.98 \ A$.
Since the galvanometer and shunt are in parallel,the potential difference across them is the same: $V = I_g \cdot r_g = I_S \cdot R_S$.
Therefore,$R_S = (I_g \cdot r_g) / I_S = 0.50 \ V / 1.98 \ A \approx 0.2525 \ \Omega$.
The value closest to this is $0.25 \ \Omega$.

(A) The ideal voltage across the $4 \text{ k}\Omega$ resistor (without the voltmeter) is calculated using the voltage divider rule: $V = \frac{4 \text{ k}\Omega}{4 \text{ k}\Omega + 4 \text{ k}\Omega} \times 4 \text{ V} = 2 \text{ V}$.
When the voltmeter with resistance $R_v = 12 \text{ k}\Omega$ is connected in parallel with the $4 \text{ k}\Omega$ resistor, the equivalent resistance $R_p$ of the parallel combination is: $R_p = \frac{4 \text{ k}\Omega \times 12 \text{ k}\Omega}{4 \text{ k}\Omega + 12 \text{ k}\Omega} = \frac{48}{16} \text{ k}\Omega = 3 \text{ k}\Omega$.
The actual reading $V^{\prime}$ of the voltmeter is the voltage across this parallel combination: $V^{\prime} = \frac{R_p}{R_p + 4 \text{ k}\Omega} \times 4 \text{ V} = \frac{3}{3 + 4} \times 4 \text{ V} = \frac{12}{7} \text{ V} \approx 1.714 \text{ V}$.
The percentage error is given by: $\text{Percentage Error} = \frac{V - V^{\prime}}{V} \times 100 = \frac{2 - \frac{12}{7}}{2} \times 100 = \frac{\frac{14 - 12}{7}}{2} \times 100 = \frac{2}{14} \times 100 = \frac{1}{7} \times 100 \approx 14.28 \%$.
Thus, the percentage error is nearly $14 \%$.

(D) The current sensitivity $I_s = 10 \text{ divisions/mA}$.
The voltage sensitivity $V_s = 2 \text{ divisions/mV}$.
The resistance of the galvanometer $G$ is given by $G = \frac{I_s}{V_s} = \frac{10 \text{ div/mA}}{2 \text{ div/mV}} = 5 \ \Omega$.
The full-scale deflection current $I_g$ for $100$ divisions is $I_g = \frac{100 \text{ divisions}}{10 \text{ divisions/mA}} = 10 \text{ mA} = 0.01 \text{ A}$.
We want the galvanometer to read $1 \text{ V}$ per division. For $100$ divisions,the total voltage $V$ required is $V = 100 \times 1 \text{ V} = 100 \text{ V}$.
Let the series resistance be $R$. Then $V = I_g(G + R)$.
$100 = 0.01(5 + R)$.
$10000 = 5 + R$.
$R = 10000 - 5 = 9995 \ \Omega$.

(B) The current sensitivity of a moving coil galvanometer is given by the formula: $S_i = \frac{\theta}{I} = \frac{NBA}{k}$.
Here,$N$ is the number of turns,$B$ is the magnetic field,$A$ is the area of the coil,and $k$ is the torsion constant of the spring.
To increase the sensitivity $(S_i)$:
$1$. Increase $N$ (number of turns).
$2$. Increase $B$ (using a stronger magnet).
$3$. Increase $A$ (using a larger coil).
$4$. Decrease $k$ (using a spring with a smaller torsion constant).
Comparing these with the given options,the $1^{st}$ way (smaller torsion constant) and the $3^{rd}$ way (stronger magnet) will increase the sensitivity. Therefore,the correct option is $B$.

(C) Let $G = 120\ \Omega$ be the resistance of the galvanometer and $S = 1\ \Omega$ be the resistance of the shunt.
Let $I_g$ be the current through the galvanometer for full-scale deflection and $I$ be the total current in the circuit.
When the shunt is connected,the total current $I = 5.5\ A$ is divided between the galvanometer and the shunt.
The current through the galvanometer is given by $I_g = I \left( \frac{S}{G + S} \right)$.
Substituting the values: $I_g = 5.5 \times \left( \frac{1}{120 + 1} \right) = 5.5 \times \left( \frac{1}{121} \right)$.
$I_g = \frac{5.5}{121} = \frac{55}{1210} = \frac{1}{22} \approx 0.04545\ A$.
Thus,the current required for full-scale deflection in the absence of the shunt is approximately $0.045\ A$.

(B) The deflection $\phi$ of a moving coil galvanometer is given by the formula: $\phi = \frac{N A B i}{k}$.
Here, $N = 50$, $r = 1 \ cm = 0.01 \ m$, $B = 0.010 \ T$, $i = 1.0 \ mA = 1.0 \times 10^{-3} \ A$, and $k = 3 \times 10^{-7} \ N-m-rad^{-1}$.
The area $A = \pi r^2 = \pi \times (0.01)^2 = \pi \times 10^{-4} \ m^2$.
Substituting these values into the formula:
$\phi = \frac{50 \times (\pi \times 10^{-4}) \times 0.010 \times 1.0 \times 10^{-3}}{3 \times 10^{-7}}$
$\phi = \frac{50 \times \pi \times 10^{-4} \times 10^{-2} \times 10^{-3}}{3 \times 10^{-7}} = \frac{50 \times \pi \times 10^{-9}}{3 \times 10^{-7}} = \frac{50 \times \pi}{3} \times 10^{-2} = \frac{0.5 \pi}{3} \approx 0.5236 \ radians$.
Converting radians to degrees: $\phi_{deg} = 0.5236 \times \frac{180}{\pi} = 30^{\circ}$.

(C) The potential difference across the galvanometer and the shunt resistor is the same because they are connected in parallel.
$V_g = V_s$
$I_g R_g = I_s R_s$
Given: $R_g = 50\,\Omega$,$R_s = 0.01\,\Omega$,$I_g = 20\,mA = 0.02\,A$.
$I_s = \frac{I_g R_g}{R_s} = \frac{0.02 \times 50}{0.01} = \frac{1}{0.01} = 100\,A$.
The total range of the ammeter $I = I_g + I_s$.
$I = 0.02\,A + 100\,A = 100.02\,A$.
Since $I_g$ is very small compared to $I_s$,the range is approximately $100\,A$.

(A) The torque on the coil is given by $\tau = NIAB \sin \theta$. For small deflections,$\tau = NIAB = C \theta$,where $C$ is the torsional constant.
Given that for current $i$,the deflection is $\theta = 90^o = \frac{\pi}{2}$ radians.
So,$NiAB = C \frac{\pi}{2} \Rightarrow C = \frac{2NiAB}{\pi}$ ........$(1)$
When a charge $Q$ is passed suddenly,the impulse of the torque is equal to the change in angular momentum:
$\int \tau dt = \int (NIAB) dt = NAB \int i dt = NABQ$
Since $\int \tau dt = \Delta L = I \omega$,we have $I \omega = NABQ$ ........$(2)$
By the law of conservation of energy,the initial kinetic energy is converted into potential energy of the spring at maximum deflection $\theta_{max}$:
$\frac{1}{2} I \omega^2 = \frac{1}{2} C \theta_{max}^2$
$\theta_{max} = \omega \sqrt{\frac{I}{C}} = \frac{I \omega}{\sqrt{IC}} = \frac{NABQ}{\sqrt{IC}}$
Substituting $C$ from $(1)$:
$\theta_{max} = \frac{NABQ}{\sqrt{I \cdot \frac{2NiAB}{\pi}}} = NABQ \sqrt{\frac{\pi}{2NiABI}} = Q \sqrt{\frac{\pi NAB}{2iI}}$

(D) To convert a galvanometer into an ammeter,a shunt resistance $S$ must be connected in parallel with the galvanometer.
Given: Galvanometer resistance $R_g = 30\,\Omega$,full-scale current $I_g = 2\,A$,and desired range $I = 10\,A$.
The formula for shunt resistance is $S = \frac{I_g R_g}{I - I_g}$.
Substituting the values: $S = \frac{2 \times 30}{10 - 2} = \frac{60}{8} = 7.5\,\Omega$.
Thus,a resistance of $7.5\,\Omega$ must be connected in parallel.

(B) The sensitivity of a galvanometer is given by $S = \frac{i_g}{i}$.
Given that the sensitivity is decreased by $30$ times,we have $\frac{i_g}{i} = \frac{1}{30}$,which implies $\frac{i}{i_g} = 30$.
The formula for the shunt resistance $S$ required to convert a galvanometer of resistance $G$ into an ammeter is $S = \frac{G \cdot i_g}{i - i_g}$.
Dividing the numerator and denominator by $i_g$,we get $S = \frac{G}{\frac{i}{i_g} - 1}$.
Substituting the given values $G = 406\, \Omega$ and $\frac{i}{i_g} = 30$:
$S = \frac{406}{30 - 1} = \frac{406}{29} = 14\, \Omega$.

(A) The shunt resistance $S$ required to convert a galvanometer of resistance $G$ into an ammeter of range $I$ is given by the formula: $S = \frac{I_g G}{I - I_g}$.
For the first range $0-I$,the shunt resistance is $S_1 = \frac{I_g G}{I - I_g}$.
For the second range $0-8I$,the shunt resistance is $S_2 = \frac{I_g G}{8I - I_g}$.
Taking the ratio of $S_1$ to $S_2$:
$\frac{S_1}{S_2} = \frac{\frac{I_g G}{I - I_g}}{\frac{I_g G}{8I - I_g}} = \frac{8I - I_g}{I - I_g}$.

(B) 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 concave (cylindrically cut) magnetic pole pieces and placing a soft iron core inside the coil.
The concave shape of the pole pieces ensures that the magnetic field lines are always parallel to the plane of the coil for any angular position,making the field radial.
Therefore,the correct option is $B$.

(D) Let $G = 36\,\Omega$ be the resistance of the galvanometer and $S = 4\,\Omega$ be the resistance of the shunt.
When a shunt is connected in parallel to a galvanometer,the current $I$ divides into two parts: $I_g$ through the galvanometer and $I_s$ through the shunt.
Since they are in parallel,the potential difference across both is the same: $I_g G = I_s S$.
Therefore,the ratio of currents is $\frac{I_s}{I_g} = \frac{G}{S} = \frac{36}{4} = 9$.
This means $I_s = 9 I_g$.
The total current is $I = I_s + I_g = 9 I_g + I_g = 10 I_g$.
The fraction of current passing through the shunt is $\frac{I_s}{I} = \frac{9 I_g}{10 I_g} = \frac{9}{10}$.

(C) The resistance of the galvanometer is $G = 50\,\Omega$.
Given that the galvanometer requires $2\,\mu A$ for $2$ divisions,the current per division is $1\,\mu A/\text{division}$.
Since there are $30$ divisions on one side,the full-scale deflection current $I_g$ is:
$I_g = 30 \times 1\,\mu A = 30 \times 10^{-6}\,A$.
The range of the ammeter is $I = 5\,A$.
The shunt resistance $S$ required is given by the formula:
$S = \frac{I_g G}{I - I_g}$.
Substituting the values:
$S = \frac{30 \times 10^{-6} \times 50}{5 - 30 \times 10^{-6}}$.
Since $30 \times 10^{-6}$ is negligible compared to $5$,we have:
$S \approx \frac{30 \times 10^{-6} \times 50}{5} = 30 \times 10^{-6} \times 10 = 300 \times 10^{-6}\,\Omega = 3 \times 10^{-4}\,\Omega$.

(A) To convert a galvanometer into an ammeter,a low resistance,known as a shunt $(S)$,must be connected in parallel with the galvanometer coil.
This arrangement ensures that the majority of the current in the circuit passes through the shunt,protecting the sensitive galvanometer coil from damage due to high current.
Additionally,it reduces the overall resistance of the device,allowing it to measure current accurately without significantly affecting the circuit's current flow.

(B) Let the main current be $I$.
According to the problem,the current passing through the galvanometer $(I_g)$ is $20\%$ of $I$,so $I_g = 0.2I$.
The current passing through the shunt $(I_s)$ is the remaining current,so $I_s = I - 0.2I = 0.8I$.
Since the galvanometer and the shunt are connected in parallel,the potential difference across them is equal:
$I_g \cdot G = I_s \cdot S$
Substituting the values:
$0.2I \cdot G = 0.8I \cdot S$
$0.2G = 0.8S$
$S = \frac{0.2G}{0.8} = \frac{G}{4}$

(B) Given: Battery voltage $E = 6\,V$,high resistance $R = 11\,k\Omega = 11000\,\Omega$,figure of merit $k = 60\,\mu A/\text{div}$,initial deflection $\theta = 9\,\text{div}$.
$1$. Calculate the current $I$ for full deflection $\theta = 9$:
$I = k \cdot \theta = 60 \times 10^{-6} \times 9 = 540 \times 10^{-6} = 5.4 \times 10^{-4}\,A$.
$2$. Using Ohm's law for the circuit without shunt:
$I = \frac{E}{R + G} \implies 5.4 \times 10^{-4} = \frac{6}{11000 + G}$.
Since $R \gg G$,$R + G \approx R = 11000\,\Omega$.
$G = \frac{E}{I} - R = \frac{6}{5.4 \times 10^{-4}} - 11000 = 11111 - 11000 = 111.1\,\Omega$.
$3$. In the half-deflection method,the shunt resistance $S$ is connected such that the deflection becomes $\theta/2 = 4.5\,\text{div}$.
The formula for shunt resistance in the half-deflection method is $S = \frac{G \cdot R}{R - G}$.
Given $R = 11000\,\Omega$ and $G \approx 111.1\,\Omega$:
$S = \frac{111.1 \times 11000}{11000 - 111.1} \approx \frac{111.1 \times 11000}{10888.9} \approx 112.2\,\Omega$.
Rounding to the nearest given option,the value is $110\,\Omega$.

(D) Given:
Galvanometer resistance,$R_g = 25\,\Omega$
Full-scale deflection current,$I_g = 1\,mA = 10^{-3}\,A$
Maximum current to be measured,$I = 2\,A$
To convert a galvanometer into an ammeter,a shunt resistance $S$ is connected in parallel with the galvanometer.
The potential difference across the galvanometer and the shunt must be equal:
$I_g R_g = (I - I_g) S$
Substituting the values:
$10^{-3} \times 25 = (2 - 10^{-3}) S$
$0.025 = (2 - 0.001) S$
$0.025 = 1.999 S$
$S = \frac{0.025}{1.999} \approx \frac{0.025}{2} = 0.0125\,\Omega$
Thus,$S = 1.25 \times 10^{-2}\,\Omega$.

(A) In the first case,the current through the galvanometer is $I_g = \frac{V}{R + G}$.
In the second case,when a shunt resistance $S$ is connected in parallel with the galvanometer,the equivalent resistance of the parallel combination is $R_p = \frac{GS}{G + S}$.
The total current in the circuit becomes $I = \frac{V}{R + R_p} = \frac{V}{R + \frac{GS}{G + S}}$.
The current through the galvanometer is now $I_g' = I \times \frac{S}{G + S} = \frac{V}{R + \frac{GS}{G + S}} \times \frac{S}{G + S} = \frac{VS}{R(G + S) + GS}$.
According to the problem,the deflection is halved,so $I_g' = \frac{I_g}{2}$.
Substituting the expressions: $\frac{VS}{R(G + S) + GS} = \frac{1}{2} \times \frac{V}{R + G}$.
$\frac{S}{RG + RS + GS} = \frac{1}{2(R + G)}$.
$2S(R + G) = RG + RS + GS$.
$2SR + 2SG = RG + RS + GS$.
$2SR - RS + 2SG - GS = RG$.
$SR + SG = RG$.
$S(R + G) = RG$.

(D) The current in the circuit without the ammeter is $I = \frac{V}{R} = \frac{5\,V}{50\,\Omega} = 0.1\,A$.
The measured current $I'$ must be within $1\%$ of $I$. Thus,$I' \geq 0.99 \times 0.1 = 0.099\,A$.
To measure current,an ammeter (galvanometer with a shunt resistor $r_s$ in parallel) is connected in series with the $50\,\Omega$ resistor.
The equivalent resistance of the circuit is $R_{eq} = 50 + R_A$,where $R_A = \frac{100 \times r_s}{100 + r_s}$.
The current measured is $I' = \frac{V}{R_{eq}} = \frac{5}{50 + R_A} = 0.099\,A$.
Solving for $R_A$: $50 + R_A = \frac{5}{0.099} \approx 50.505\,\Omega$.
So,$R_A = 50.505 - 50 = 0.505\,\Omega$.
Since $R_A = \frac{100 \times r_s}{100 + r_s} = 0.505$,we get $100 r_s = 50.5 + 0.505 r_s$.
$99.495 r_s = 50.5 \Rightarrow r_s \approx 0.507\,\Omega$.
Rounding to the nearest provided option,$r_s = 0.5\,\Omega$ connected in parallel is the correct choice.

(D) Let $I$ be the full scale deflection current and $V = 2\,V$ be the battery voltage.
In case $1$,when $R_1 = 2400\,\Omega$,deflection $\theta_1 = 40$ divisions.
$\frac{40}{50} I = \frac{V}{G + R_1} \Rightarrow \frac{4}{5} I = \frac{2}{G + 2400} \dots (1)$
In case $2$,when $R_2 = 4900\,\Omega$,deflection $\theta_2 = 20$ divisions.
$\frac{20}{50} I = \frac{V}{G + R_2} \Rightarrow \frac{2}{5} I = \frac{2}{G + 4900} \dots (2)$
Dividing $(1)$ by $(2)$:
$\frac{4/5 I}{2/5 I} = \frac{G + 4900}{G + 2400} \Rightarrow 2 = \frac{G + 4900}{G + 2400}$
$2G + 4800 = G + 4900 \Rightarrow G = 100\,\Omega$.
Substituting $G = 100\,\Omega$ in $(1)$:
$\frac{4}{5} I = \frac{2}{100 + 2400} = \frac{2}{2500} = \frac{1}{1250}$
$I = \frac{5}{4} \times \frac{1}{1250} = \frac{1}{1000}\,A = 1\,mA$.
Current sensitivity $= \frac{I}{50} = \frac{1\,mA}{50} = 0.02\,mA/\text{division} = 20\,\mu A/\text{division}$.
For a deflection of $10$ divisions:
$\frac{10}{50} I = \frac{V}{G + R} \Rightarrow \frac{1}{5} \times 10^{-3} = \frac{2}{100 + R}$
$100 + R = 10000 \Rightarrow R = 9900\,\Omega$.
Thus,option $D$ is correct.

(D) In the half-deflection method,a high resistance $R$ is connected in series with the galvanometer to maintain a nearly constant current,and a shunt resistance $S$ is connected in parallel with the galvanometer to reduce the deflection to half.
The circuit diagram $D$ shows a high resistance $R$ in series with the galvanometer $G$,and a shunt resistance $S$ connected in parallel with the galvanometer through a key $K_2$.
When key $K_1$ is closed and $K_2$ is open,the current $I$ flows through $R$ and $G$. When $K_2$ is closed,the shunt $S$ is introduced,and the current through the galvanometer becomes $I/2$. The galvanometer resistance $G$ is then given by the formula $G = \frac{RS}{R - S}$.

(D) Let $G = 120\,\Omega$ be the resistance of the galvanometer and $S = 1\,\Omega$ be the resistance of the shunt.
Let $I = 5.5\,A$ be the total current flowing through the parallel combination.
Let $I_g$ be the current flowing through the galvanometer for full-scale deflection.
Using the current divider rule for a parallel circuit,the current through the galvanometer is given by:
$I_g = I \times \frac{S}{G + S}$
Substituting the given values:
$I_g = 5.5 \times \frac{1}{120 + 1}$
$I_g = 5.5 \times \frac{1}{121}$
$I_g = \frac{5.5}{121} = \frac{55}{1210} = \frac{1}{22} \approx 0.04545\,A$
Therefore,the current that gives full-scale deflection is approximately $0.045\,A$.