Converting a Galvanometer to an Ammeter and a Voltmeter

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Electric Circuits Intermediate Ohm's Law

Source: High school physics (Chinese)

Problem Sets:

ohm's law

Problem

A galvanometer has an internal resistance $R_g = 90 \ \Omega$ and a full-scale deflection current $I_g = 0.02 \ \text{A}$.

  1. If a $10 \ \Omega$ shunt resistor is connected in parallel with the galvanometer, what is the full-scale range of the resulting ammeter?
  2. If a $360 \ \Omega$ multiplier resistor is connected in series with the galvanometer, what is the full-scale range of the resulting voltmeter?

Ammeter range: $0.2 \ \text{A}$. Voltmeter range: $9 \ \text{V}$.

\textbf{Ammeter (shunt $R_s = 10 \ \Omega$ in parallel).} The shunt carries the excess current $I - I_g$, with the same voltage as the galvanometer:

$$I_g R_g = (I - I_g) R_s \implies I = I_g\!\left(1 + \dfrac{R_g}{R_s}\right) = 0.02 \times (1 + 9) = 0.2 \ \text{A}.$$

\textbf{Voltmeter (multiplier $R_m = 360 \ \Omega$ in series).} At full scale the same current $I_g$ flows through both:

$$U = I_g (R_g + R_m) = 0.02 \times (90 + 360) = 9 \ \text{V}.$$