Correct Answer is Option (C)

Current sensitivity of moving coil galvanometer

I_s = $\frac{NBA}{C}$ .....(i)

Voltage sensitivity of moving coil galvanometer,

V_s = $\frac{NBA}{C{R}_G}$

Dividing eqn. (i) by (ii)

Resistance of galvanometer

R_G = $\frac{I_s}{V_S}=\frac{5\times 1}{20\times {10}^{-3}}=\frac{5000}{20}=250\mathrm{\Omega }$

Correct Answer is Option (C)

In parallel arrangement, I $\alpha$ R

Now G/S = 99.8/0.2

S = G/499

Hence, $R_A=\frac{GS}{G+S}=\frac{G}{500}$

Correct Answer is Option (A)

$S=\frac{V_g}{\left(I-I_g\right)}$

Neglecting I_g

$\therefore$ $S=\frac{V_g}{I}=\frac{25\times {10}^{-3}V}{25A}=0.001\mathrm{\Omega }$

Correct Answer is Option (D)

To keep the main current in the circuit unchanged, the resistance of the galvanometer should be equal to the net resistance.

$\therefore$ $G=\left(\frac{GS}{G+S}\right)+S^{\prime }$

$\Rightarrow G-\frac{GS}{G+S}=S^{\prime }$

$\therefore S^{\prime }=\frac{G^2}{G+S}$

Correct Answer is Option (A)

Here,
Resistance of galvanometer, G = 100 $\mathrm{\Omega }$

Current for full scale deflection, I_g = 30 mA = 30 × 10^–3 A

Range of voltmeter, V = 30 V

To convert the galvanometer into an voltmeter of a given range, a resistance R is connected in series with it as shown in the figure.


From figure, V = I_g(G + R)

$\Rightarrow R=\frac{V}{I_g}-G=\frac{30}{30\times {10}^{-3}}-100\mathrm{\Omega }$

= 1000 – 100 = 900 $\mathrm{\Omega }$

Correct Answer is Option (A)

Let the resistance to be added be R, then

30 = I_g (r + R)

$\therefore$ R = $\frac{30}{I_g}-r$

= $\frac{30}{30\times {10}^{-3}}-100$

= 1000 - 100 = 900 $\mathrm{\Omega }$

Correct Answer is Option (C)

Correct Answer is Option (B)

Total initial resistance

= R_G + R₁ = (50 + 2950) $\mathrm{\Omega }$ = 3000 $\mathrm{\Omega }$

$\epsilon =3V$

$\therefore$ Current = $\frac{3V}{3000\mathrm{\Omega }}=1\times {10}^{-3}mA$

If the deflection has to be reduced to 20 divisions, current i = 1 mA $\times$ $\frac{2}{3}$ as the full deflection scale for 1 mA = 30 divisions.

$3V=3000\mathrm{\Omega }\times 1mA=x\mathrm{\Omega }\times \frac{2}{3}mA$

$\Rightarrow x=3000\times 1\times \frac{3}{2}=4500\mathrm{\Omega }$

But the galvanometer resistance = 50 $\mathrm{\Omega }$

Therefore the resistance to be added

= (4500 – 50)$\mathrm{\Omega }$ = 4450 $\mathrm{\Omega }$.

Correct Answer is Option (A)

We know

$\frac{I}{I_S}=1+\frac{G}{S}$

$\frac{750}{100}=1+\frac{13}{S}$

$\Rightarrow S=2\mathrm{\Omega }$

Correct Answer is Option (D)

The total current shown by the galvanometer,

I_g = 25 $\times$ 4 $\times$ 10^-4 A = 10^-2 A

The value of resistance connected in series to convert galvanometer into voltmeter of 25 V is

V = I_g(R_e + R_g)

$\Rightarrow$ R_e = $\frac{V}{I_g}-R_g$ = 2450 $\mathrm{\Omega }$

Correct Answer is Option (A)

For converting galvanometer to voltmeter, a high resistance should be connected in series with galvanometer.