Respuesta :
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'.
⇒ [tex]V=20\ \Omega,\ V_1=30\ \Omega[/tex]
⇒ [tex]G=1680\ \Omega,\ G_1=2930\ \Omega[/tex]
Concept to be used:
Conversion of galvanometer into voltmeter.
Let [tex]G[/tex] be the resistance of the galvanometer and [tex]I_g[/tex] the maximum deflection in the galvanometer.
To measure maximum voltage resistance [tex]R[/tex] is connected in series .
So,
⇒ [tex]V=I_g(R+G)[/tex]
We have to find the value of [tex]R[/tex] we know that in series circuit current are same.
For [tex]G=1680[/tex] For [tex]G_1=2930[/tex]
⇒ [tex]I_g=\frac{V}{R+G}[/tex] equation (i) ⇒ [tex]I_g=\frac{V_1}{R+G_1}[/tex] equation (ii)
Equating both the above equations:
⇒ [tex]\frac{V}{R+G} = \frac{V_1}{R+G_1}[/tex]
⇒ [tex]V(R+ G_1) = V_1 (R+G)[/tex]
⇒ [tex]VR+VG_1 = V_1R+V_1G[/tex]
⇒ [tex]VR-V_1R = V_1G-VG_1[/tex]
⇒ [tex]R(V-V_1) = V_1G-VG_1[/tex]
⇒ [tex]R =\frac{V_1G-VG_1}{(V-V_1)}[/tex]
⇒ Plugging the values.
⇒ [tex]R =\frac{(30\times 1680) - (20\times 2930)}{(20-30)}[/tex]
⇒ [tex]R =\frac{(50400 - 58600)}{(-10)}[/tex]
⇒ [tex]R=\frac{-8200}{-10}[/tex]
⇒ [tex]R=820\ \Omega[/tex]
The coil resistance of the circuit is 820 Ω .
