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Chapter 10: Problem 59

What voltage is involved in a \(1.44-\mathrm{kW}\) short circuit through a \(0.100-\Omega\) resistance?

Short Answer

Expert verified
The voltage involved in a 1.44 kW short circuit through a \(0.100\,\Omega\) resistance is \(3.8\,\text{V}\).

Step by step solution

01

Find the current using power dissipation formula

The power dissipation formula for a resistor is given by: \[P = I^2R\] Where, P = Power dissipation (1.44 kW) I = Current flowing through the resistor R = Resistance (0.1惟) Given P = 1.44 kW, which is equivalent to 1440 W, we can rearrange the formula to find I: \[I = \sqrt{\frac{P}{R}}\]
02

Calculate the current

Now, we can plug the values of P and R into the formula to find the current. \[I = \sqrt{\frac{1440}{0.1}}\] By calculating the square root, we will find the current I: \[I = 38\,\text{A}\]
03

Calculate the voltage using Ohm's law

Now that we know the current, we can use Ohm's law to find the voltage across the resistor. Ohm's law states that: \[V = IR\] Plugging in the values for I and R, we get: \[V = (38\,\text{A})(0.1\,\Omega)\]
04

Find the voltage

Multiplying the current and resistance values, we get the voltage across the resistor: \[V = 3.8\,\text{V}\] So, the voltage involved in a 1.44 kW short circuit through a 0.1惟 resistance is 3.8 V.

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Key Concepts

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

Understanding Power Dissipation
Power dissipation in electrical circuits is essentially the loss of electrical power, usually in the form of heat. It's crucial because it helps us assess how much energy is being converted in a component like a resistor. In our example, we dealt with a power of 1.44 kW, which we first convert to watts, resulting in 1440 W. This conversion is necessary because standard formulas work with watts rather than kilowatts.

To determine how power relates to a resistor, we use the formula for power dissipation:
  • Power ( P ) = Current ( I ) ^ 2 multiplied by the resistance ( R ) .
This equation allows us to find how much current is flowing through a resistor, considering the power value given.
Exploring the Role of the Resistor
A resistor is a commonly used component in electrical circuits designed to resist the flow of current. Its key function is to control the amount of current passing through a circuit. Resistors can also dissipate power, often in the form of heat, as they limit electrical flow. In our specific scenario, the resistor has a low value of 0.1 惟, which means it offers minimal resistance to the current flow, allowing high amounts of current.

Understanding a resistor's role includes knowing:
  • Resistance value ( R ) dictates how much it restricts current flow.
  • Power dissipation indicates the heat energy produced as current flows through.
With these principles, you can calculate other essential variables like current and voltage, enhancing circuit stability and performance.
Steps in Current Calculation
The calculation of current is significant to find out how much electrical flow moves through a resistor. In our exercise, we used the power dissipation formula to determine the current. We rearrange the formula to isolate the current (I), achieving: \[I = \sqrt{\frac{P}{R}}\]
This formula highlights:
  • The proportional relationship between powerand the square of the current.
  • The inverse relationship between resistanceand current.
Plugging the given values, 1440 W for power and 0.1 惟 for resistance, gives:\[I = \sqrt{\frac{1440}{0.1}} = 38\,\text{A}\]Thus, 38 A of current flows through the resistor. This understanding is critical for designing and managing electrical circuits efficiently.
Voltage Calculation through Ohm's Law
Voltage is the electrical potential difference across a component, like a resistor. It's computed using Ohm's Law, which links voltage (V), current (I), and resistance (R) in a simple linear relationship: \[V = IR\]
Once the current is known, voltage calculation becomes direct. We got the current as 38 A from the previous calculation. Using the given resistance of the resistor (0.1 惟), we apply Ohm's Law to find voltage: \[V = (38\,\text{A}) \times (0.1\,\Omega) = 3.8\,\text{V}\]This means the voltage involved in such a short circuit is 3.8 V. This use of Ohm's Law demonstrates how understanding current and resistance helps in finding voltage accurately in circuits.

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Most popular questions from this chapter

A flashing lamp in a Christmas earring is based on an \(R C\) discharge of a capacitor through its resistance. The effective duration of the flash is 0.250 s, during which it produces an average 0.500 W from an average 3.00 V. (a) What energy does it dissipate? (b) How much charge moves through the lamp? (c) Find the capacitance. (d) What is the resistance of the lamp? (since average values are given for some quantities, the shape of the pulse profile is not needed.)

(a) During surgery, a current as small as \(20.0 \mu \mathrm{A}\) applied directly to the heart may cause ventricular fibrillation. If the resistance of the exposed heart is \(300 \Omega,\) what is the smallest voltage that poses this danger? (b) Does your answer imply that special electrical safety precautions are needed?

What are the advantages and disadvantages of connecting batteries in series? In parallel?

A heart defibrillator passes 10.0 A through a patient's torso for \(5.00 \mathrm{ms}\) in an attempt to restore normal beating. (a) How much charge passed? (b) What voltage was applied if \(500 \mathrm{J}\) of energy was dissipated? (c) What was the path's resistance? (d) Find the temperature increase caused in the \(8.00 \mathrm{kg}\) of affected tissue.

A \(120-\mathrm{V}\) immersion heater consists of a coil of wire that is placed in a cup to boil the water. The heater can boil one cup of \(20.00^{\circ} \mathrm{C}\) water in 180.00 seconds. You buy one to use in your dorm room, but you are worried that you will overload the circuit and trip the \(15.00-\mathrm{A}\) 120-V circuit breaker, which supplies your dorm room. In your dorm room, you have four \(100.00-\mathrm{W}\) incandescent lamps and a \(1500.00-\mathrm{W}\) space heater. (a) What is the power rating of the immersion heater? (b) Will it trip the breaker when everything is turned on? (c) If it you replace the incandescent bulbs with 18.00-W LED, will the breaker trip when everything is turned on?
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