Question

practical applications

you are an electrician on the job. you have been given a multimeter that has the following ac voltage ranges: 30, 60, and 150. the meter states that it has a resistance of 5000 \\(\omega\\)/v. you need to be able to measure a voltage of 277 volts. how much resistance should be inserted in series with the meter to make the 30-volt range indicate a full-scale value of 300 volts?

Explanation:

Identify given values

We are given:

• Original range: \(V_{\text{old}} = 30\text{ V}\)
• Sensitivity: \(S = 5000\ \Omega/\text{V}\)
• Target full-scale voltage: \(V_{\text{new}} = 300\text{ V}\)

Calculate original meter resistance

Using Voltmeter Sensitivity, the total internal resistance of the meter on the \(30\text{-V}\) range is:

$$ R_{\text{meter}} = V_{\text{old}} \times S $$

$$ R_{\text{meter}} = 30\text{ V} \times 5000\ \Omega/\text{V} = 150,000\ \Omega $$

Calculate total required resistance

Using Voltmeter Sensitivity and Ohms Law, the total resistance required to measure up to \(300\text{ V}\) is:

$$ R_{\text{total}} = V_{\text{new}} \times S $$

$$ R_{\text{total}} = 300\text{ V} \times 5000\ \Omega/\text{V} = 1,500,000\ \Omega $$

Calculate series multiplier resistance

The additional series resistance \(R_{\text{series}}\) needed is:

$$ R_{\text{series}} = R_{\text{total}} - R_{\text{meter}} $$

$$ R_{\text{series}} = 1,500,000\ \Omega - 150,000\ \Omega = 1,350,000\ \Omega $$

$$ R_{\text{series}} = 1.35\text{ M}\Omega $$

Answer:

\(1,350,000\ \Omega\) (or \(1.35\text{ M}\Omega\))