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Chapter 26: Problem 39

A circuit consists of two \(1.00-\mathrm{k} \Omega\) resistors in series with an ideal \(12.0-\mathrm{V}\) battery. a) Calculate the current flowing through each resistor. b) A student trying to measure the current flowing through one of the resistors inadvertently connects an ammeter in parallel with that resistor rather than in series with it. How much current will flow through the ammeter, assuming that it has an internal resistance of \(1.0 \Omega ?\)

Short Answer

Expert verified
Answer: The current flowing through each resistor in the series circuit is \(6.00\,\mathrm{mA}\). When the ammeter is connected in parallel to one of the resistors, the current flowing through the ammeter is approximately \(12.0\,\mathrm{mA}\).

Step by step solution

01

Calculate the equivalent resistance of the circuit

To calculate the current flowing through each resistor, we first need to find the equivalent resistance of the circuit. Since the two resistors are in series, we can find the equivalent resistance by simply adding their values: \(R_{eq} = R_1 + R_2 = 1.00\,\mathrm{k\Omega} + 1.00\,\mathrm{k\Omega} = 2.00\,\mathrm{k\Omega}\)
02

Calculate current using Ohm's Law

Now that we have the equivalent resistance, we can use Ohm's Law to find the current flowing through the circuit. Ohm's Law states that \(V = IR,\) where V is the voltage, I is the current, and R is the resistance. In our case, we have: \(I = \dfrac{V}{R_{eq}} = \dfrac{12.0\,\mathrm{V}}{2.00\,\mathrm{k\Omega}} = 6.00\,\mathrm{mA}\) Since the resistors are in series, the current flowing through each resistor is the same, which is \(6.00\,\mathrm{mA}.\)
03

Calculate the equivalent resistance of the circuit with the ammeter

When the student connects the ammeter in parallel with one of the resistors, the equivalent resistance of the parallel combination of the resistor and the ammeter is: \(R_{par} = \dfrac{R_3 \cdot R_A}{R_3 + R_A} = \dfrac{1.00\,\mathrm{k\Omega} \cdot 1.0\,\mathrm{\Omega}}{1.00\,\mathrm{k\Omega} + 1.0\,\mathrm{\Omega}} \approx 1.0\,\mathrm{\Omega}\)
04

Calculate the new equivalent resistance of the circuit

Now, we have the old resistor replaced by the equivalent parallel resistance (including the ammeter). So, we can calculate the new equivalent resistance of the circuit by adding the second resistor and the equivalent parallel resistance: \(R_{eq2} = R_2 + R_{par} = 1.00\,\mathrm{k\Omega} + 1.0\,\mathrm{\Omega} \approx 1.00\,\mathrm{k\Omega}\)
05

Calculate the current using Ohm's Law with the new equivalent resistance

Now that we have the new equivalent resistance, we can use Ohm's Law again to find the current flowing through the circuit: \(I_2 = \dfrac{V}{R_{eq2}} = \dfrac{12.0\,\mathrm{V}}{1.00\,\mathrm{k\Omega}} = 12.0\,\mathrm{mA}\)
06

Calculate the current flowing through the ammeter

Since the ammeter is connected in parallel with one of the resistors, we need to find the current flowing through the ammeter using the current divider rule: \(I_A = I_2 \dfrac{R_3}{R_3 + R_A} = 12.0\,\mathrm{mA} \times \dfrac{1.00\,\mathrm{k\Omega}}{1.00\,\mathrm{k\Omega} + 1.0\,\mathrm{\Omega}} \approx 12.0\,\mathrm{mA}\) So, the current flowing through the ammeter is approximately \(12.0\,\mathrm{mA}\).

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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 Wheatstone bridge is constructed using a \(1.00-\mathrm{m}-\) long Nichrome wire (the purple line in the figure) with a conducting contact that can slide along the wire. A resistor, \(R_{1}=\) \(100 . \Omega\), is placed on one side of the bridge, and another resistor, \(R,\) of unknown resistance, is placed on the other side. The contact is moved along the Nichrome wire, and it is found that the ammeter reading is zero for \(L=25.0 \mathrm{~cm} .\) Knowing that the wire has a uniform cross section throughout its length, determine the unknown resistance.

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.)

Consider a series \(\mathrm{RC}\) circuit with \(R=10.0 \Omega\) \(C=10.0 \mu \mathrm{F}\) and \(V=10.0 \mathrm{~V}\) a) How much time, expressed as a multiple of the time constant, does it take for the capacitor to be charged to half of its maximum value? b) At this instant, what is the ratio of the energy stored in the capacitor to its maximum possible value? c) Now suppose the capacitor is fully charged. At time \(t=\) 0 , the original circuit is opened and the capacitor is allowed to discharge across another resistor, \(R^{\prime}=1.00 \Omega\), that is connected across the capacitor. What is the time constant for the discharging of the capacitor? d) How many seconds does it take for the capacitor to discharge half of its maximum stored charge, \(Q\) ?

An ammeter with an internal resistance of \(53 \Omega\) measures a current of \(5.25 \mathrm{~mA}\) in a circuit containing a battery and a total resistance of \(1130 \Omega\). The insertion of the ammeter alters the resistance of the circuit, and thus the measurement does not give the actual value of the current in the circuit without the ammeter. Determine the actual value of the current.
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