Question

Two scales on a nondigital voltmeter measure voltages up to 20.0 and 30.0 V, respectively. The resistance connected in series with the galvanometer is 1680 Ω for the 20.0-V scale and 2930 Ω for the 30.0-V scale. Determine the coil resistance.

Answer 1

Answer:

Resistance of the circuit is 820 Ω

Explanation:

Given:

Two galvanometer resistance are given along with its voltages.

Let the resistance is "R" and the values of voltages be 'V' and 'V1' along with 'G' and 'G1'.

⇒ $V=20\ \Omega,\ V_1=30\ \Omega$

⇒ $G=1680\ \Omega,\ G_1=2930\ \Omega$

Concept to be used:

Conversion of galvanometer into voltmeter.

Let $G$ be the resistance of the galvanometer and $I_g$ the maximum deflection in the galvanometer.

To measure maximum voltage resistance $R$ is connected in series .

So,

⇒ $V=I_g(R+G)$

We have to find the value of $R$ we know that in series circuit current are same.

For $G=1680$                                    For $G_1=2930$

⇒ $I_g=\frac{V}{R+G}$   equation (i)                ⇒ $I_g=\frac{V_1}{R+G_1}$ equation (ii)

Equating both the above equations:

⇒ $\frac{V}{R+G} = \frac{V_1}{R+G_1}$

⇒ $V(R+ G_1) = V_1 (R+G)$

⇒ $VR+VG_1 = V_1R+V_1G$

⇒ $VR-V_1R = V_1G-VG_1$

⇒ $R(V-V_1) = V_1G-VG_1$

⇒ $R =\frac{V_1G-VG_1}{(V-V_1)}$

⇒ Plugging the values.

⇒ $R =\frac{(30\times 1680) - (20\times 2930)}{(20-30)}$

⇒ $R =\frac{(50400 - 58600)}{(-10)}$

⇒ $R=\frac{-8200}{-10}$

⇒ $R=820\ \Omega$

The coil resistance of the circuit is 820 Ω .