Current Electricity
Akopọ
Current electricity is a fundamental concept in physics that deals with the flow of electric charge in a circuit. In this course, we will delve into various aspects of current electricity, focusing on key topics such as electromagnetic force (emf), potential difference (p.d.), current, internal resistance of a cell, and lost Volt.
One of the primary objectives of this course is to differentiate between electromagnetic force, potential difference, current, and internal resistance of a cell. Understanding these concepts is crucial as they form the basis of electrical circuits and their behavior. By grasping the differences between these terms, students will be able to analyze circuit parameters effectively.
Another key objective is to apply Ohm’s law to solve problems related to current electricity. Ohm’s law states that the current flowing through a conductor is directly proportional to the potential difference across it, provided the temperature remains constant. By mastering Ohm’s law, students will be equipped to calculate unknown electrical quantities in circuits.
The course also covers the measurement of resistance using techniques such as the meter bridge. The meter bridge is a useful tool that allows for precise determination of resistance in a circuit. By learning how to use the meter bridge, students can accurately measure resistance and understand its significance in circuit analysis.
Furthermore, students will explore the concepts of resistance in series and in parallel, as well as their combinations. Understanding how resistances behave in series and parallel configurations is essential for designing and analyzing complex circuits. By studying these configurations, students will gain insights into optimizing circuit performance.
Moreover, the course will introduce students to the potentiometer method of measuring emf, current, and internal resistance of a cell. The potentiometer is a versatile instrument that offers high precision in measuring electrical quantities. By utilizing the potentiometer, students can accurately measure key parameters in a circuit.
Lastly, the course will delve into electrical networks and the application of Kirchoff’s law. Kirchoff’s laws, including Kirchoff's voltage law and Kirchoff's current law, are fundamental principles in circuit analysis. By applying these laws, students can solve complex network problems and understand the behavior of current in circuits.
Awọn Afojusun
- Determine The Resistivity And The Conductivity Of A Conductor
- Differentiate Between Emf, Pd, Current And Internal Resistance Of A Cell
- Measure Emf, Current And Internal Resistance Of A Cell Using The Potentiometer
- Use Metre Bridge To Calculate Resistance
- Apply Kirchoff’s Law In Electrical Networks
- Identify The Advantages Of The Potentiometer
- Apply Ohm’s Law To Solve Problems
Akọ̀wé Ẹ̀kọ́
Forúkọsílẹ̀ láti Tẹ̀síwájú
Ṣẹda àkọọlẹ ọfẹ kan láti wọlé sí gbogbo àwọn oríṣìíríṣìí ìkànsí ikẹ́kọ̀ọ́, àwọn ìbéèrè ìdánwò, àti láti tọpa ìlọsíwájú rẹ.
Ìdánwò Ẹ̀kọ́
Oriire fun ipari ẹkọ lori Current Electricity. Ni bayi ti o ti ṣawari naa awọn imọran bọtini ati awọn imọran, o to akoko lati fi imọ rẹ si idanwo. Ẹka yii nfunni ni ọpọlọpọ awọn adaṣe awọn ibeere ti a ṣe lati fun oye rẹ lokun ati ṣe iranlọwọ fun ọ lati ṣe iwọn oye ohun elo naa.
Iwọ yoo pade adalu awọn iru ibeere, pẹlu awọn ibeere olumulo pupọ, awọn ibeere idahun kukuru, ati awọn ibeere iwe kikọ. Gbogbo ibeere kọọkan ni a ṣe pẹlu iṣaro lati ṣe ayẹwo awọn ẹya oriṣiriṣi ti imọ rẹ ati awọn ogbon ironu pataki.
Lo ise abala yii gege bi anfaani lati mu oye re lori koko-ọrọ naa lagbara ati lati ṣe idanimọ eyikeyi agbegbe ti o le nilo afikun ikẹkọ. Maṣe jẹ ki awọn italaya eyikeyi ti o ba pade da ọ lójú; dipo, wo wọn gẹgẹ bi awọn anfaani fun idagbasoke ati ilọsiwaju.
- What is the formula for calculating potential difference (p.d.) in a circuit? A. V = IR B. V = I/R C. V = I x R D. V = R/I Answer: B. V = I/R
- What is the SI unit of current? A. Volts B. Watts C. Amperes D. Ohms Answer: C. Amperes
- Which of the following best describes Ohm's Law? A. Resistance is directly proportional to voltage B. Current is inversely proportional to resistance C. Current is directly proportional to voltage D. Voltage remains constant in a circuit Answer: C. Current is directly proportional to voltage
- What is the function of the internal resistance of a cell in a circuit? A. To generate electrical energy B. To control the flow of current C. To increase the potential difference D. To provide stability to the circuit Answer: B. To control the flow of current
- In which configuration does the total resistance in a circuit decrease - series or parallel? A. Series B. Parallel Answer: B. Parallel
- Which instrument is commonly used to measure resistance accurately? A. Ammeter B. Voltmeter C. Ohmmeter D. Galvanometer Answer: C. Ohmmeter
- What is the principle behind the operation of a potentiometer in measuring emf? A. Kirchoff's law B. Joule's law C. Principle of moments D. Wheatstone bridge principle Answer: D. Wheatstone bridge principle
- How is the conductivity of a conductor related to its resistivity? A. They are inversely proportional B. They are directly proportional C. They are not related D. Conductivity depends on temperature only Answer: B. They are directly proportional
- Which law is used to analyze complex electrical networks? A. Faraday's law B. Newton's law C. Kirchoff's law D. Ohm's law Answer: C. Kirchoff's law
Forúkọsílẹ̀ láti Tẹ̀síwájú
Ṣẹda àkọọlẹ ọfẹ kan láti wọlé sí gbogbo àwọn oríṣìíríṣìí ìkànsí ikẹ́kọ̀ọ́, àwọn ìbéèrè ìdánwò, àti láti tọpa ìlọsíwájú rẹ.
Forúkọsílẹ̀ láti Tẹ̀síwájú
Ṣẹda àkọọlẹ ọfẹ kan láti wọlé sí gbogbo àwọn oríṣìíríṣìí ìkànsí ikẹ́kọ̀ọ́, àwọn ìbéèrè ìdánwò, àti láti tọpa ìlọsíwájú rẹ.
Awọn Iwe Itọsọna Ti a Gba Nimọran
Ibeere Atunyewo
Ṣe o n ronu ohun ti awọn ibeere atijọ fun koko-ọrọ yii dabi? Eyi ni nọmba awọn ibeere nipa Current Electricity lati awọn ọdun ti o kọja.
Ibeere 1 Ìròyìn
You are provided with a battery of e.m.f, E, a standard resistor, R, of resistance 2 \(\Omega\), a key, K, an ammeter, A, a jockey, J, a potentiometer, UV, and some connecting wires.
(i) Measure and record the emf, E, of the battery.
(ii) Set up the circuit as shown in the diagram above with the key open.
(iii) Place the jockey at the point, U, of the potentiometer wire. Close the key and record the reading, i, of the ammeter.
(iv) Place the jockey at a point T on the potentiometer wire UV such that d = UT = 30.0 cm.
(v) Close the circuit, read and record the current, I, on the ammeter,
(vi) Evaluate \(I^1\).
(vi) Repeat the experiment for four other values of d = 40.0 cm, 50.0 cm, 60.0 cm and 70.0 cm. In each case, record I and evaluate \(I^1\).
(vii) Tabulate the results
(ix) Plot a graph with d on the vertical axis and I on the horizontal axis stalling both axes from the origin (0,0).
(x) Determine the slope, s, of the graph.
(xi) From the graph determine the value \(I_1\), of I when d = 0. (ci) Given that=s, calculate 8.
(xii) State two precautions taken to ensure accurate results.
(xii) Given that \(\frac{E}{\delta}\) = s, calculate \(\delta\).
(b)(i) Write down the equation that connects the resistance, R, of a wire and the factors on which it depends. State the meaning of each of the symbols.
(ii) An electric fan draws a current of0.75 A in a 240 V circuit. Calculate the cost of using, the fan for 10 hours if the utility rate is $ 0.50 per kWh.
Awọn alaye Idahun
(a) Potentiometer experiment
(i) The e.m.f. of the battery is:
\[E=2.0\ \text{V}\]
With the jockey at \(U\), the ammeter reading is:
\[i=1.00\ \text{A}\]
(ii) Table of results
| Distance, \(d\) (cm) | Current, \(I\) (A) | \(I^{-1}\) (A−1) |
|---|---|---|
| 30.0 | 0.769 | 1.30 |
| 40.0 | 0.714 | 1.40 |
| 50.0 | 0.667 | 1.50 |
| 60.0 | 0.625 | 1.60 |
| 70.0 | 0.588 | 1.70 |
As \(d\) increases, \(I\) decreases.
(iii) Graph of \(d\) against \(I^{-1}\)
(iv) Slope of the graph
Using two widely separated points on the straight line, \((1.20\ \text{A}^{-1},20.0\ \text{cm})\) and \((1.70\ \text{A}^{-1},70.0\ \text{cm})\):
\[s=\frac{70.0-20.0}{1.70-1.20}=\frac{50.0}{0.50}=100\ \text{cm A}.\]
(v) Value of \(I^{-1}\) when \(d=0\)
From the intercept on the \(I^{-1}\)-axis,
\[I^{-1}=1.00\ \text{A}^{-1}.\]
Therefore,
\[I=\frac{1}{1.00}=1.00\ \text{A},\]
which agrees with the current \(i\) when the jockey is at \(U\).
(vi) Resistance per unit length, \(\delta\)
Given that
\[s=\frac{E}{\delta},\]
\[\delta=\frac{E}{s}=\frac{2.0}{100}=0.020\ \Omega\,\text{cm}^{-1}.\]
(vii) Precautions
- The key was closed only while taking a reading, to prevent heating of the potentiometer wire and consequent change in resistance.
- The jockey was touched lightly on the wire and not pressed hard, to ensure good contact without damaging the wire.
(b)
(i) The resistance of a uniform wire is given by
\[R=\rho\frac{l}{A}.\]
\(R\) is the resistance of the wire, \(\rho\) is the resistivity of the material, \(l\) is the length of the wire, and \(A\) is its cross-sectional area.
(ii)
\[P=VI=240\times0.75=180\ \text{W}=0.180\ \text{kW}.\]
Energy used in 10 hours:
\[E=Pt=0.180\times10=1.80\ \text{kWh}.\]
Cost of energy:
\[\text{Cost}=1.80\times\$0.50=\boxed{\$0.90}.\]
Forúkọsílẹ̀ láti Tẹ̀síwájú
Ṣẹda àkọọlẹ ọfẹ kan láti wọlé sí gbogbo àwọn oríṣìíríṣìí ìkànsí ikẹ́kọ̀ọ́, àwọn ìbéèrè ìdánwò, àti láti tọpa ìlọsíwájú rẹ.
Ibeere 1 Ìròyìn
Forúkọsílẹ̀ láti Tẹ̀síwájú
Ṣẹda àkọọlẹ ọfẹ kan láti wọlé sí gbogbo àwọn oríṣìíríṣìí ìkànsí ikẹ́kọ̀ọ́, àwọn ìbéèrè ìdánwò, àti láti tọpa ìlọsíwájú rẹ.
Ibeere 1 Ìròyìn
Awọn alaye Idahun
To calculate the monthly electrical bill, we first need to determine the total energy consumption of the household in kilowatt-hours (kWh). Here are the steps:
1. Calculate the total power consumption of the appliances daily:
- The appliances' power ratings are 80W, 100W, and 120W.
- Total power consumption: 80W + 100W + 120W = 300W.
2. Convert the daily power consumption from Watts to kilowatts (kW):
- Since 1kW = 1000W, the daily power consumption in kW is 300W / 1000 = 0.3 kW.
3. Calculate the energy used daily in kWh:
- The appliances are used for 6 hours daily.
- Energy used daily: 0.3 kW * 6 hours = 1.8 kWh.
4. Calculate the monthly energy consumption:
- An average month is 31 days.
- Monthly energy consumption: 1.8 kWh/day * 31 days = 55.8 kWh.
5. Calculate the cost based on the rate:
- The cost of electricity is ₦24.08 per kWh.
- Monthly electricity cost: 55.8 kWh * ₦24.08 = ₦1343.664.
Therefore, the monthly electrical bill is approximately ₦1343.66k.
Forúkọsílẹ̀ láti Tẹ̀síwájú
Ṣẹda àkọọlẹ ọfẹ kan láti wọlé sí gbogbo àwọn oríṣìíríṣìí ìkànsí ikẹ́kọ̀ọ́, àwọn ìbéèrè ìdánwò, àti láti tọpa ìlọsíwájú rẹ.