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Chapter 26: Problem 55

The ammeter your physics instructor uses for in-class demonstrations has internal resistance \(R_{\mathrm{i}}=75 \Omega\) and measures a maximum current of \(1.5 \mathrm{~mA}\). The same ammeter can be used to measure currents of much greater magnitudes by wiring a shunt resistor of relatively small resistance, \(R_{\text {shunt }}\), in parallel with the ammeter. (a) Sketch the circuit diagram, and explain why the shunt resistor connected in parallel with the ammeter allows it to measure larger currents. (b) Calculate the resistance the shunt resistor has to have to allow the ammeter to measure a maximum current of 15 A.

Short Answer

Expert verified
Question: Sketch a circuit diagram with an ammeter and a shunt resistor connected in parallel and explain how this setup allows the ammeter to measure larger currents. Calculate the resistance of the shunt resistor that must be used to measure a maximum current of 15 A, given that the ammeter has an internal resistance of 75 Ohms and a maximum current of 1.5 mA. Answer: When an ammeter and a shunt resistor are connected in parallel, the current gets divided between them based on their respective resistances. The ammeter measures the current flowing through it and, by knowing the resistance values of both the ammeter and the shunt resistor, we can calculate the total current flowing through the circuit. To measure a maximum current of 15 A, a shunt resistor with a resistance of approximately 0.0075 Ohms is required.

Step by step solution

01

Sketch the circuit diagram with the ammeter and shunt resistor connected in parallel

First, draw a simple circuit with a battery, a resistor representing the device we want to measure the current of (load), and the ammeter connected in series. Next, add the shunt resistor in parallel with the ammeter, which means connecting one end of the shunt resistor to the same point where the ammeter connects to the circuit, and the other end to the same point where the ammeter connects to the load. This creates a parallel connection between the ammeter and the shunt resistor.
02

Understand the parallel connection of the ammeter and shunt resistor

When the ammeter and shunt resistor are connected in parallel, the current flowing through the circuit is divided between the two resistors according to their resistance. The ammeter measures the current flowing through it, and since we know the resistance of the ammeter and the shunt resistor, we can determine the total current flowing through the circuit. In this way, the ammeter can measure larger currents that would otherwise be beyond its maximum limit. The combined resistance of the ammeter and shunt resistor effectively becomes smaller, allowing more current to flow.
03

Write down equations to determine the shunt resistor's resistance

We know the formula for the equivalent resistance of parallel resistors is: \(\frac{1}{R_{total}} = \frac{1}{R_{i}} + \frac{1}{R_{shunt}}\) We also know the maximum current the ammeter can measure (1.5 mA) and the maximum current we want it to measure (15 A). Let's denote the current passing through the ammeter as I_i and the current passing through the shunt resistor as I_shunt. We can write down the following equations: \(I_{total} = I_{i} + I_{shunt}\) \(I_i = \frac{V}{R_i}\) \(I_{shunt} = \frac{V}{R_{shunt}}\) We want to find the value of R_shunt that enables the ammeter to measure a maximum current of 15 A.
04

Solving for R_shunt when the maximum current is 15 A

First, we calculate the maximum voltage drop across the internal resistance of the ammeter: \(V_{max} = R_i \times I_{max} = 75 \Omega \times 1.5 \times 10^{-3} A = 0.1125 V\) To measure a total current of 15 A, we need the current through the shunt resistor to be: \(I_{shunt} = I_{total} - I_i = 15 A - 1.5 \times 10^{-3} A = 14.9985 A\) Now we can calculate the required shunt resistance: \(R_{shunt} = \frac{V_{max}}{I_{shunt}} = \frac{0.1125 V}{14.9985 A} = 0.0075 \Omega\) The shunt resistor must have a resistance of approximately 0.0075 Ohms to allow the ammeter to measure a maximum current of 15 A.

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Most popular questions from this chapter

Many electronics devices can be dangerous even after they are shut off. Consider an RC circuit with a \(150 .-\mu \mathrm{F}\) capacitor and a \(1.00-\mathrm{M} \Omega\) resistor connected to a \(200 .-\mathrm{V}\) power source for a long time and then disconnected and shorted, as shown in the figure. How long will it be until the potential difference across the capacitor drops to below \(50.0 \mathrm{~V} ?\)

If the capacitor in an \(\mathrm{RC}\) circuit is replaced with two identical capacitors connected in series, what happens to the time constant for the circuit?

Two capacitors in series are charged through a resistor. Identical capacitors are instead connected in parallel and charged through the same resistor. How do the times required to fully charge the two sets of capacitors compare?

Explain why the time constant for an \(\mathrm{RC}\) circuit increases with \(R\) and with \(C\). (The answer "That's what the formula says" is not sufficient.)

A battery, a resistor, and a capacitor are connected in series in an RC circuit. What happens to the current through a resistor after a long time? Explain using Kirchhoff's rules.
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