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Chapter 9: Problem 35

Calculate the effective resistance of a pocket calculator that has a 1.35 -V battery and through which \(0.200 \mathrm{mA}\) flows.

Short Answer

Expert verified
The effective resistance of the pocket calculator is \(6750 \,\Omega\).

Step by step solution

01

Convert the current to Amperes

We are given the current value as 0.200 mA (milliamperes). To convert it to Amperes (A), we need to divide it by 1000: \[ I = \frac{0.200 \,\mathrm{mA}}{1000} = 0.0002 \,\mathrm{A} \]
02

Use Ohm's Law to find the effective resistance

Plug the values of V (voltage) and I (current in Amperes) into Ohm's Law formula to find the effective resistance R: \[R =\frac{V}{I} \] Now, we can substitute the values of V and I in the equation: \[R = \frac{1.35 \,\mathrm{V}}{0.0002 \,\mathrm{A}} \]
03

Solve for R

Divide 1.35 V by 0.0002 A to get the value of R: \[R = 6750 \,\Omega \] So, the effective resistance of the pocket calculator is 6750 Ohms.

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Key Concepts

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

Effective Resistance
When we talk about effective resistance, we're referring to the total resistance faced by electric current in a circuit. It's the combined resistance of all the components in the circuit. In a simple device like a pocket calculator, there's essentially a single pathway for the current to flow, and the effective resistance is just the resistance of that path.

In the provided exercise, the calculation of effective resistance is done by applying Ohm's Law. The process requires the voltage across the device and the current flowing through it. By dividing the voltage by the current, we obtain the effective resistance. It's imperative to note that all components in the circuit contribute to this value and any alteration in these components would change the effective resistance, leading to a different current flow for the same voltage.
Electric Current Conversion
Understanding electric current conversion is essential when working with electrical units. Electric current, which is the flow of electric charge, is measured in amperes (A). Often, current is given in milliamperes (mA), especially when dealing with small electronic devices like calculators.

To convert milliamperes to amperes, we divide the current in milliamperes by 1,000 since one ampere is equal to one thousand milliamperes. This conversion is crucial because Ohm's Law, which is used to calculate effective resistance, requires that current be in amperes. Mishandling this conversion can lead to incorrect calculations and results, so attention to detail is paramount.
Voltage-Current Relation
The voltage-current relation is a fundamental concept in electrical circuits, defined by Ohm's Law. The law states that the current (I) through a conductor between two points is directly proportional to the voltage (V) across the two points and inversely proportional to the resistance (R) of the conductor. Expressed as an equation, it is: \[ R = \frac{V}{I} \]

This relationship is crucial for understanding how a change in voltage or resistance affects the current flow in a circuit. In the context of a pocket calculator, by knowing the voltage provided by the battery and the current flowing, we can deduce the effective resistance of the calculator. This principle also helps in designing circuits with the desired electrical characteristics and for troubleshooting issues when a device is not operating as intended.

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Most popular questions from this chapter

A particle accelerator produces a beam with a radius of \(1.25 \mathrm{mm}\) with a current of \(2.00 \mathrm{mA}\). Each proton has a kinetic energy of \(10.00 \mathrm{MeV}\). (a) What is the velocity of the protons? (b) What is the number (n) of protons per unit volume? (b) How many electrons pass a cross sectional area each second?

What requirement for superconductivity makes current superconducting devices expensive to operate?

The current \(I\) is measured through a sample of an ohmic material as a voltage \(V\) is applied. (a) What is the current when the voltage is doubled to \(2 \mathrm{V}\) (assume the change in temperature of the material is negligible)? (b) What is the voltage applied is the current measured is \(0.2 I\) (assume the change in temperature of the material is negligible)? What will happen to the current if the material if the voltage remains constant, but the temperature of the material increases significantly?

Consider a wire of a circular cross-section with a radius of \(R=3.00 \mathrm{mm}\). The magnitude of the current density is modeled as \(J=c r^{2}=5.00 \times 10^{6} \frac{\mathrm{A}}{\mathrm{m}^{4}} r^{2} .\) What is the current through the inner section of the wire from the center to \(r=0.5 R ?\)

Resistors are commonly rated at \(\frac{1}{8} \mathrm{W}, \frac{1}{4} \mathrm{W}, \frac{1}{2} \mathrm{W}\) 1 W and 2 W for use in electrical circuits. If a current of \(I=2.00 \mathrm{A}\) is accidentally passed through a \(R=1.00 \Omega\) resistor rated at \(1 \mathrm{W}\), what would be the most probable outcome? Is there anything that can be done to prevent such an accident?
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