Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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.

Short Answer

Expert verified
Answer: The actual value of the current in the circuit without the ammeter is approximately 5.44 mA.

Step by step solution

01

Calculate the original resistance of the circuit without the ammeter

Without the ammeter, the total resistance of the circuit is \(R = 1130 \Omega\). We want to find the actual current without the ammeter.
02

Calculate the equivalent resistance of the circuit with the ammeter

The internal resistance of the ammeter is \(R_A = 53 \Omega\). When the ammeter is inserted into the circuit, the equivalent resistance of the circuit becomes the sum of the original resistance and the ammeter's resistance: \(R_{eq} = R + R_A\).
03

Calculate the actual current in the circuit

The ammeter measures a current of \(5.25 \mathrm{mA}\) in the circuit with an altered resistance of \(R_{eq}\). Let \(I_{measured}\) be the measured current and \(I_{actual}\) be the actual current we want to find. We can apply Ohm's Law (V=IR) to both the circuit with and without the ammeter: \(I_{actual} R = I_{measured} R_{eq}\) Now, we need to solve for \(I_{actual}\). Since we know the value of the resistances and the measured current, we can plug in values: \(I_{actual} = \frac{I_{measured} R_{eq}}{R}\)
04

Plug in values and solve for the actual current

We can now plug in the values for the measured current and resistances into the formula: \(I_{actual} = \frac{(5.25 \mathrm{~mA})(1130 \Omega + 53 \Omega)}{1130 \Omega}\) \(I_{actual} = \frac{5.25 \mathrm{~mA} \cdot 1183 \Omega}{1130 \Omega}\) \(I_{actual} \approx 5.44 \mathrm{~mA}\) Therefore, the actual value of the current in the circuit without the ammeter is approximately \(5.44 \mathrm{~mA}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ammeter
An ammeter is a crucial instrument used in electrical circuits to measure the current flowing through the circuit. It is connected in series to the circuit, meaning it becomes a part of the circuit's electric path. This positioning allows it to directly measure the current. However, every ammeter has an internal resistance, which can affect the circuit’s total resistance when it is inserted.

The internal resistance of an ammeter can cause a slight deviation in the actual current measurement. For example, in the given problem, the ammeter’s internal resistance was 53 Ohms. This resistance affected the circuit's total resistance and, consequently, the measured current. To ensure accurate readings, it's crucial to consider an ammeter's internal resistance in some precise measurements or calculations.
Resistance
Resistance is the opposition that a material offers to the flow of electric current. It is measured in ohms (Ω). In circuits, resistance affects how much current will flow for a given voltage, as explained by Ohm's Law (V=IR).

In the problem at hand, the circuit had a given total resistance of 1130 Ω. When the ammeter, with its internal resistance of 53 Ω, was added to the circuit, the total resistance increased. This new total resistance had to be considered to find the accurate current flowing through the circuit.
  • Internal Resistance: Causes the total resistance to increase when an ammeter is added.
  • Circuit Resistance: A sum of all resistive components in a given path.
Understanding resistance is essential for analyzing circuits and predicting how they will behave under various conditions.
Measured Current
Measured current is the value that an ammeter displays when it is part of a circuit. This current can differ from the actual current due to the added resistance from the ammeter. In the mentioned problem, the measured current was 5.25 mA.

Due to the additional resistance introduced by the ammeter, the measured current was less than the actual current in the circuit without the ammeter. To determine the actual current, we first find the total resistance with the ammeter and then use Ohm's Law to correct the measurement. It's important to differentiate between measured current and actual current to ensure electrical devices function efficiently and safely.
Circuit Analysis
Circuit analysis is the process of determining the currents and voltages in a circuit. This includes understanding how different components and their characteristics, like resistance, impact the overall circuit behavior.

There are several methods to analyze circuits. Ohm’s Law is one fundamental tool used in these analyses. In the given problem, we used Ohm's Law to compare the circuit's behavior with and without the ammeter. This involved knowing the equivalence of circuit resistances and using these values to calculate the actual current.
  • Ohm's Law Application: Formulating equations to relate currents with resistances and voltages.
  • Series and Parallel Configurations: Affect how components influence total resistance.
Circuit analysis ensures we can calculate accurately and predictively manage components in an electrical circuit to work efficiently.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A circuit consists of two \(100 .-\mathrm{k} \Omega\) resistors in series with an ideal \(12.0-\mathrm{V}\) battery. a) Calculate the potential drop across one of the resistors. b) A voltmeter with internal resistance \(10.0 \mathrm{M} \Omega\) is connected in parallel with one of the two resistors in order to measure the potential drop across the resistor. By what percentage will the voltmeter reading deviate from the value you determined in part (a)? (Hint: The difference is rather small so it is helpful to solve algebraically first to avoid a rounding error.)

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.

Kirchhoff's Junction Rule states that a) the algebraic sum of the currents at any junction in a circuit must be zero. b) the algebraic sum of the potential changes around any closed loop in a circuit must be zero. c) the current in a circuit with a resistor and a capacitor varies exponentially with time. d) the current at a junction is given by the product of the resistance and the capacitance. e) the time for the current development at a junction is given by the product of the resistance and the capacitance.

Two parallel plate capacitors, \(C_{1}\) and \(C_{2},\) are con nected in series with a \(60.0-\mathrm{V}\) battery and a \(300 .-\mathrm{k} \Omega\) resistor, as shown in the figure. Both capacitors have plates with an area of \(2.00 \mathrm{~cm}^{2}\) and a separation of \(0.100 \mathrm{~mm}\). Capacitor \(C_{1}\) has air between its plates, and capacitor \(C_{2}\) has the gap filled with a certain porcelain (dielec-

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.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free