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Chapter 12: Problem 8

When a 12 V battery is connected across an unknown resistor, there is a current of \(2.5 \mathrm{~mA}\) in the circuit. Find the value of the resistance of the resistor.

Short Answer

Expert verified
The resistance of the resistor is 4800 ohms.

Step by step solution

01

- Identify given values

Identify the values provided in the exercise. The voltage across the resistor, V, is 12 V and the current through the resistor, I, is 2.5 mA or 0.0025 A.
02

- Use Ohm's Law

Write down Ohm's Law, which is given by the formula: \[ V = I \times R \]Where V is voltage, I is current, and R is resistance.
03

- Rearrange Ohm's Law to solve for resistance

To find the resistance, rearrange the formula to solve for R: \[ R = \frac{V}{I} \]
04

- Substitute the given values into the rearranged formula

Substitute the known values into the equation: \[ R = \frac{12 \text{ V}}{0.0025 \text{ A}} \]
05

- Calculate the resistance

Carry out the division to find the value of the resistance: \[ R = 4800 \text{ } \text{ohms} \]

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Key Concepts

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

Understanding Voltage
Voltage is a fundamental concept in electronics and electrical engineering. It is the potential difference between two points in a circuit. Think of it like the pressure in a water pipe. The higher the voltage, the more force that pushes electrons through a conductor.

Voltage is measured in volts (V). In our example, the battery provides a voltage of 12 V. This means there is a potential difference of 12 volts between the battery's terminals, which drives the current through the resistor.

Here are some key points to remember:
  • Voltage is often denoted by the letter V.
  • It's the driving force that moves electrical charge through a circuit.
  • Higher voltage means more energy is available to move electrons.
Understanding voltage is vital for grasping how electrical circuits work.
Exploring Current
Current is the flow of electric charge in a circuit. Imagine it as the water flowing through a pipe due to pressure. The amount of current in a circuit depends on both the voltage and the resistance.

Current is measured in amperes (A), but in many small circuits, you might encounter milliamperes (mA), where 1 A = 1000 mA. In our example, we have a current of 2.5 mA, which can also be written as 0.0025 A.

Key points about current include:
  • Current is usually denoted by the letter I.
  • It's the rate at which charge flows through a point in the circuit.
  • Ohm's Law helps us understand the relationship between current, voltage, and resistance.
Remember, for a given voltage, higher resistance means lower current, and vice versa.
Resistance Calculation
Resistance is a measure of how much a component opposes the flow of current. Think of it like the diameter of a pipe: a small diameter resists water flow more than a wide one.

Resistance is measured in ohms (Ω). Ohm's Law provides a straightforward way to calculate resistance if we know the current and voltage. The formula is \( R = \frac{V}{I} \).

Using the values from our example, we can find the resistance:
\[ R = \frac{12 \text{ V}}{0.0025 \text{ A}} = 4800 \text { } \text{ohms} \]

Key points about resistance include:
  • Resistance is denoted by R.
  • It's calculated using the ratio of voltage to current.
  • Higher resistance reduces current for a given voltage.
Remember, knowing how to calculate resistance helps you design and troubleshoot circuits effectively.

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Most popular questions from this chapter

A battery of \(9 \mathrm{~V}\) is connected in series with resistors of \(0.2 \Omega, 0.3 \Omega, 0.4 \Omega, 0.5 \Omega\) and \(12 \Omega\), respectively. How much current would flow through the \(12 \Omega\) resistor?

Several electric bulbs designed to be used on a \(220 \mathrm{~V}\) electric supply line, are rated \(10 \mathrm{~W}\). How many lamps can be connected in parallel with each other across the two wires of \(220 \mathrm{~V}\) line if the maximum allowable current is \(5 \mathrm{~A}\) ?

Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be- (a) \(1: 2\) (b) \(2: 1\) (c) \(1: 4\) (d) \(4: 1\)

Show how you would connect three resistors, each of resistance \(6 \Omega\), so that the combination has a resistance of (i) \(9 \Omega\), (ii) \(4 \Omega\).

A piece of wire of resistance \(R\) is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is \(R^{\prime}\), then the ratio \(R / R^{\prime}\) is - (a) \(1 / 25\) (b) \(1 / 5\) (c) 5 (d) 25
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