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Chapter 26: Problem 38

You want to make an ohmmeter to measure the resistance of unknown resistors. You have a battery with voltage \(\mathrm{V}_{\mathrm{emf}}=9.00 \mathrm{~V}\), a variable resistor, \(R,\) and an ammeter that measures current on a linear scale from 0 to \(10.0 \mathrm{~mA}\) a) What resistance should the variable resistor have so that the ammeter gives its full-scale (maximum) reading when the ohmmeter is shorted? b) Using the resistance from part (a), what is the unknown resistance if the ammeter reads \(\frac{1}{4}\) of its full scale?

Short Answer

Expert verified
Answer: The variable resistor should have a resistance of 900 Ω for the ammeter to show its maximum reading. The unknown resistor's resistance is 2700 Ω when the ammeter reads one-quarter of its full scale.

Step by step solution

01

Ohm's Law

First, let's recall Ohm's Law: \(V = IR\) where V is voltage, I is current, and R is resistance.
02

Find the variable resistor's resistance for a maximum ammeter reading

For part (a), we want to find the resistance of the variable resistor when the ammeter reads its maximum value, which is 10.0 mA. Since the circuit is shorted, there is no other resistance present, and we can use Ohm's Law. First, convert the maximum current to amperes: \(I = 10.0 \times 10^{-3} A = 0.010 A.\) Now, apply Ohm's Law to find the resistance: \(R = \frac{V_{emf}}{I} = \frac{9.00 V}{0.010 A} = 900 \Omega\) The variable resistor should have a resistance of 900 Ω such that the ammeter shows its maximum reading when the circuit is shorted.
03

Find the unknown resistance when the ammeter reads one-quarter of its full scale

For part (b), when the ammeter reads one-quarter of its full scale, the current passing through the circuit can be calculated as follows: \(I = \frac{1}{4}(10.0 \times 10^{-3} A) = 0.00250 A.\) Let's now use Ohm's Law again, this time to find the voltage across the unknown resistor: \(V_{unknown} = V_{emf} - V_{variable} = V_{emf} - IR_{variable} = 9.00 V - (0.00250 A)(900 \Omega) = 6.75 V\) Now, we can find the unknown resistor's resistance, \(R_{unknown}\), by applying Ohm's Law once more: \(R_{unknown} = \frac{V_{unknown}}{I} = \frac{6.75 V}{0.00250 A} = 2700 \Omega\) So the unknown resistor has a resistance of 2700 Ω when the ammeter reads one-quarter of its full scale.

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Most popular questions from this chapter

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.

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.

How long would it take, in multiples of the time constant, \(\tau,\) for the capacitor in an \(\mathrm{RC}\) circuit to be \(98 \%\) charged? a) \(9 \tau\) c) \(90 \tau\) e) \(0.98 \tau\) b) \(0.9 \tau\) d) \(4 \tau\)

A capacitor bank is designed to discharge 5.0 J of energy through a \(10.0-\mathrm{k} \Omega\) resistor array in under \(2.0 \mathrm{~ms}\) To what potential difference must the bank be charged, and what must the capacitance of the bank be?

A circuit consists of a source of emf, a resistor, and a capacitor, all connected in series. The capacitor is fully charged. How much current is flowing through it? a) \(i=V / R\) b) zero c) neither (a) nor (b)
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