Solving Electrical Circuit Homework Questions

In summary, when there is a potential difference of 1.0 V between two points in an electrical ciruit, the electron loses a mean amount of energy (the electronvolt) when moving between these two points. This amount of energy has a special name, the electronvolt, symbol eV. When an electron is accelerated across a potential difference of about 15 V, it has enough energy to ionise a nitrogen atom. This process is called ionisation and it happens with the release of a small amount of energy, called an attojoule.
  • #1
kitkat794
1
0

Homework Statement



i've been trying to do these but i keep getting different answers from back of book.

1)When there is a potential difference of 1.0 V between two points in an electrical ciruit, what is the mean amount of energy lost by an electron when moving between these two points? This amount of energy has a special name, the electronvolt, symbol eV.

2)When an electron is accelerated across a potential difference of about 15 V it has enough energy to ionise a nitrogen atom, that is, to remove an electron from the atom by collision with it. How many attojoule, aJ is this? 1.0aJ=1.0 x 10^-18 J.

3) The electrical circuit in a small torch consists of a battery of e.m.f 3.0 V and a bulb in series. The potential difference across the bulb is 2.7 V when the current in the bulb is 0.3 A.

a) Calcuate how much energy is converted in one minute:
i) from chemical form to electrica form in battery
ii) from electrical form to thermal energy and light in the bulb



Homework Equations







The Attempt at a Solution



1) ?
2) 1 volt= 1 joule per couloumb second so 15V = 15 joules percoloumb second. then i did 15 divided by 1x10^-18 and i got 1.5 x10^19 aJ which is wrong (should be 2.4aJ)

3)a) i know i meant to use the formula for electromotive force but i don't know how to work out the charge to put it in this (answer should be 54J)

ii) use formua for potential difference which again needs charge and i don't know how to work it out (answer should be 48.6 J)
 
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  • #2
1. energy = charge * potential diff.

multiply charge of electron with 1V, that is 1 eV
 
  • #3
2. same case, energy is 15V*(charge of 1 electron), convert units.
 
  • #4
For 3, do you know how to determine the power (rate of energy delivered or consumed) given voltage and current? Then, if you have the power, how can you find the total energy over a given time period?
 
  • #5


I would first like to commend you for attempting to solve these electrical circuit homework questions. It is important to remember that in science, there can be multiple approaches to solving a problem and it is common to get different answers when working through a problem. However, there are also certain principles and equations that must be followed in order to arrive at the correct answer.

In regards to the first question, the mean amount of energy lost by an electron when moving between two points with a potential difference of 1.0 V is indeed called an electronvolt (eV). This is equivalent to 1.6 x 10^-19 joules. Therefore, in order to find the energy lost by an electron when moving between two points with a potential difference of 1.0 V, you would simply multiply 1.6 x 10^-19 by 1.0 V, resulting in 1.6 x 10^-19 joules.

For the second question, it is important to remember that 1 volt is equal to 1 joule per coulomb. Therefore, when an electron is accelerated across a potential difference of 15 V, it has 15 joules of energy per coulomb. To find the energy in attojoules (aJ), you would simply multiply 15 by 1.0 x 10^-18, resulting in 1.5 x 10^-17 aJ.

Moving on to the third question, it is important to use the equation P = IV, where P is power, I is current, and V is potential difference. In this case, we are given the current (0.3 A) and potential difference (2.7 V) for the bulb, so we can calculate the power as 0.3 A x 2.7 V = 0.81 watts. To find the energy converted in one minute, we simply multiply 0.81 watts by 60 seconds, resulting in 48.6 J.

For part (ii) of this question, we can use the same equation, P = IV, but this time we are given the potential difference (3.0 V) and current (0.3 A) for the battery. Therefore, the power is 0.3 A x 3.0 V = 0.9 watts. Again, to find the energy in one minute, we multiply 0.9 watts by
 

FAQ: Solving Electrical Circuit Homework Questions

What is an electrical circuit?

An electrical circuit is a path or network of conductors and components through which electricity can flow. It is used to power devices and systems by providing a continuous flow of electrical energy.

What is Ohm's Law and how is it used to solve circuit problems?

Ohm's Law is a fundamental law of electricity that states the relationship between voltage, current, and resistance in a circuit. It states that the current flowing through a conductor is directly proportional to the voltage applied and inversely proportional to the resistance of the conductor. This law is used to solve circuit problems by calculating the unknown value (voltage, current, or resistance) using the known values and the formula V=IR.

What is the difference between series and parallel circuits?

In a series circuit, all components are connected in a single loop, so the current is the same at all points. In a parallel circuit, the components are connected in multiple branches, so the current is split between the branches. Additionally, in a series circuit, the total resistance is equal to the sum of individual resistances, while in a parallel circuit, the total resistance is less than the smallest individual resistance.

How do you calculate the total resistance of a series or parallel circuit?

In a series circuit, the total resistance is equal to the sum of individual resistances. So, to calculate the total resistance, add all the individual resistances together. In a parallel circuit, the total resistance is less than the smallest individual resistance. So, to calculate the total resistance, use the formula 1/Rt = 1/R1 + 1/R2 + 1/R3 + ... + 1/Rn, where Rt is the total resistance and R1, R2, R3, etc. are the individual resistances.

What are Kirchhoff's laws and how are they applied to solve circuit problems?

Kirchhoff's laws are two fundamental laws of electricity that govern the behavior of currents and voltages in a circuit. The first law, also known as Kirchhoff's current law, states that the sum of currents entering a node (junction) in a circuit must equal the sum of currents leaving the node. The second law, also known as Kirchhoff's voltage law, states that the sum of voltage drops around a closed loop in a circuit must equal the sum of the voltage sources in that loop. These laws are applied to solve circuit problems by writing and solving sets of equations based on the laws and the given circuit diagram.

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