Question

problem 19.59
an ammeter has a sensitivity of 44000 ω/v
part b
what is the resistance of a voltmeter on the 250 - v scale if the meter sensitivity is 44000 ω/v?
express your answer to two significant figures and include the appropriate units.

Explanation:

Step1: Recall ammeter - sensitivity formula

The sensitivity of an ammeter is given by $S=\frac{R_m}{I_{fs}}$, where $S$ is the sensitivity, $R_m$ is the resistance of the meter movement, and $I_{fs}$ is the full - scale current. For a basic ammeter, when we consider the sensitivity formula in terms of the relationship between resistance and voltage (since $V = IR$), and the fact that the sensitivity $S=\frac{R}{V}$ at full - scale. For an ammeter, the full - scale current $I_{fs}$ can be found from $S=\frac{1}{I_{fs}}$ (assuming the internal resistance of the ammeter for the sensitivity formula is considered in the context of Ohm's law). So $I_{fs}=\frac{1}{S}$.
Given $S = 44000\ \Omega/V$, then $I_{fs}=\frac{1}{44000}\ A=22.73\ \mu A\approx23\ \mu A$.

Step2: Calculate voltmeter resistance

The resistance of a voltmeter $R_V$ is given by the product of the sensitivity $S$ and the full - scale voltage $V$. Given $S = 44000\ \Omega/V$ and $V = 250\ V$. Then $R_V=S\times V$.
$R_V=44000\times250\ \Omega = 1.1\times10^{7}\ \Omega$.

Answer:

Part A:
Value: $23$
Units: $\mu A$
Part B:
Value: $1.1\times 10^{7}$
Units: $\Omega$