\n\n\n\n
The movement of charge in a conductor from the positive terminal to the negative terminal is referred to as electricity. Electricity has a wide range of uses and applications. It serves as a medium for supplying power to electrical devices. We already know that the flow of charge generates current, which we call electricity.<\/p>\n\n\n\n
Electric Current Formula<\/h2>\n\n\n\n
To understand how electricity is produced, we must first understand the various basic parameters associated with it, such as voltage, current, resistance, conductivity, and the relationships between these quantities.<\/p>\n\n\n\n
Symbols and Notations<\/h2>\n\n\n\nPhysical quantities, symbols, and SI Units<\/h3>\n\n\n\n![\"Physical](\"https:\/\/physicscatalyst.com\/article\/wp-content\/uploads\/2021\/04\/class-10-electricity-notations.png\")
<\/a>Physical quantities, symbols and SI Units used in electricity formulas<\/figcaption><\/figure><\/div>\n\n\n\nSymbols for commonly used electrical components<\/h3>\n\n\n\n![\"Symbols](\"https:\/\/physicscatalyst.com\/article\/wp-content\/uploads\/2021\/04\/electric-circuit-components.png\")
<\/a>Symbols of various components used in electric circuits<\/figcaption><\/figure>\n\n\n\nFormulas of electricity<\/h2>\n\n\n\n
Given below are all the formulas used for class 10 electricity chapter.<\/p>\n\n\n\n
\n- Charge q on a body is always denoted by
\n$q = ne$
\nwhere n = any integer positive or negative and $e =1.602 \\times 10^{-19} C$ i. e., charge on an electron or proton.<\/li>\n - $Work\\; done = charge \\times potential $ or potential difference
\nMathematically,
\n$W=qV=q(V_2V_1)\\; Joule$<\/li>\n - $Electric\\; Current = \\frac{charge}{time}$
\nOr,
\n$I=\\frac{q}{t}\\;Ampere$<\/li>\n - Ohm\u2019s Law<\/strong>
\n$Resistance=\\frac{potential\\; difference}{current}$
\nOr,
\n$R=\\frac{V}{I}$<\/li>\n- Resistance in terms of resistivity<\/strong>
\n$R=\\frac{\\rho l}{A}$<\/li>\n- Resistance in series combination<\/strong>
\n$R_s=R_1+R_2+R_3+\u2026..$<\/li>\n- Resistance in parallel combination<\/strong>
\n$\\frac{1}{R_p}= \\frac{1}{R_1}+ \\frac{1}{R_2}+ \\frac{1}{R_3}+\u2026..$<\/li>\n- Electrical Power<\/strong>
\n$P=VI=I^2R=\\frac{V^2}{R}$<\/li>\n- Power in Series<\/strong>
\n$\\frac{1}{P_s}= \\frac{1}{P_1}+ \\frac{1}{P_2}+ \\frac{1}{P_3}+\u2026..$<\/li>\n- Power in parallel<\/strong>
\n$P_p=P_1+P_2+P_3+\u2026..$<\/li>\n- Electrical Energy<\/strong>
\n$E-Vit=I^2Rt=\\frac{V^2}{R}t$<\/li>\n- Heat produced<\/strong>
\n$H-Vit=I^2Rt=\\frac{V^2}{R}t$<\/li>\n<\/ol>\n\n\n\nSolved Examples <\/h2>\n\n\n\n
Question 1 <\/strong>
When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.
Solution
Here,
Battery Voltage $V= 12 V$
Current in circuit $I=2.5mA=2.5\\times 10^{-3}A$
Resistance of the resistor $R=?$ is to be calculated
We know that from Ohm\u2019s Law
$R=\\frac{V}{I}$
Putting in the values we get
$R=\\frac{12V}{2.5\\times 10^{-3}A}$
$=4.8\\times 10^{3}VA^{-1}=4.8\\times 10^{3}\\Omega $<\/p>\n\n\n\nQuestion 2<\/strong>
An electric heater of resistance $8\\Omega$ draws 15 A from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater.
Solution
It is given in the question that
Heater resistance $R=8\\Omega$
Current drawn, $I=15 A$
Time for which current is drawn
$t=2h=2\\times 60\\times 60 \\;sec=7200\\; sec$
Rate of heat developed $P=?$ which is to be calculated
We know that
Rate of heat developed = rate at which power is consumed.
$P=I^2R$
Putting in the value we get,
$P=(15 A)^2\\times 8\\omega=1800W$
The rate of heat developed is
$P=1800 J\/s$ Test Yourself on questions based on these formulas<\/p>\n\n\n\nTest Yourself <\/h2>\n\n\n\n
\n\n\n\n
General Instructions<\/strong><\/p>\n\n\n\n- Your test contains multiple-choice questions with only one answer type of questions. There are a total of 5 questions<\/li>
- This is a 10 min test. Please make sure you complete it in the stipulated time<\/li>
- You can finish this test any time using the ‘View Results<\/em><\/strong>‘ button.<\/li><\/ul>\n\n\n\n
\n\n\n
<\/a>Symbols for commonly used electrical components<\/h3>\n\n\n\n<\/a>Symbols of various components used in electric circuits<\/figcaption><\/figure>\n\n\n\nFormulas of electricity<\/h2>\n\n\n\n
Given below are all the formulas used for class 10 electricity chapter.<\/p>\n\n\n\n
\n- Charge q on a body is always denoted by
\n$q = ne$
\nwhere n = any integer positive or negative and $e =1.602 \\times 10^{-19} C$ i. e., charge on an electron or proton.<\/li>\n - $Work\\; done = charge \\times potential $ or potential difference
\nMathematically,
\n$W=qV=q(V_2V_1)\\; Joule$<\/li>\n - $Electric\\; Current = \\frac{charge}{time}$
\nOr,
\n$I=\\frac{q}{t}\\;Ampere$<\/li>\n - Ohm\u2019s Law<\/strong>
\n$Resistance=\\frac{potential\\; difference}{current}$
\nOr,
\n$R=\\frac{V}{I}$<\/li>\n- Resistance in terms of resistivity<\/strong>
\n$R=\\frac{\\rho l}{A}$<\/li>\n- Resistance in series combination<\/strong>
\n$R_s=R_1+R_2+R_3+\u2026..$<\/li>\n- Resistance in parallel combination<\/strong>
\n$\\frac{1}{R_p}= \\frac{1}{R_1}+ \\frac{1}{R_2}+ \\frac{1}{R_3}+\u2026..$<\/li>\n- Electrical Power<\/strong>
\n$P=VI=I^2R=\\frac{V^2}{R}$<\/li>\n- Power in Series<\/strong>
\n$\\frac{1}{P_s}= \\frac{1}{P_1}+ \\frac{1}{P_2}+ \\frac{1}{P_3}+\u2026..$<\/li>\n- Power in parallel<\/strong>
\n$P_p=P_1+P_2+P_3+\u2026..$<\/li>\n- Electrical Energy<\/strong>
\n$E-Vit=I^2Rt=\\frac{V^2}{R}t$<\/li>\n- Heat produced<\/strong>
\n$H-Vit=I^2Rt=\\frac{V^2}{R}t$<\/li>\n<\/ol>\n\n\n\nSolved Examples <\/h2>\n\n\n\n
Question 1 <\/strong>
When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.
Solution
Here,
Battery Voltage $V= 12 V$
Current in circuit $I=2.5mA=2.5\\times 10^{-3}A$
Resistance of the resistor $R=?$ is to be calculated
We know that from Ohm\u2019s Law
$R=\\frac{V}{I}$
Putting in the values we get
$R=\\frac{12V}{2.5\\times 10^{-3}A}$
$=4.8\\times 10^{3}VA^{-1}=4.8\\times 10^{3}\\Omega $<\/p>\n\n\n\nQuestion 2<\/strong>
An electric heater of resistance $8\\Omega$ draws 15 A from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater.
Solution
It is given in the question that
Heater resistance $R=8\\Omega$
Current drawn, $I=15 A$
Time for which current is drawn
$t=2h=2\\times 60\\times 60 \\;sec=7200\\; sec$
Rate of heat developed $P=?$ which is to be calculated
We know that
Rate of heat developed = rate at which power is consumed.
$P=I^2R$
Putting in the value we get,
$P=(15 A)^2\\times 8\\omega=1800W$
The rate of heat developed is
$P=1800 J\/s$ Test Yourself on questions based on these formulas<\/p>\n\n\n\nTest Yourself <\/h2>\n\n\n\n
\n\n\n\n
General Instructions<\/strong><\/p>\n\n\n\n- Your test contains multiple-choice questions with only one answer type of questions. There are a total of 5 questions<\/li>
- This is a 10 min test. Please make sure you complete it in the stipulated time<\/li>
- You can finish this test any time using the ‘View Results<\/em><\/strong>‘ button.<\/li><\/ul>\n\n\n\n
\n\n\n
<\/a>Formulas of electricity<\/h2>\n\n\n\n
Given below are all the formulas used for class 10 electricity chapter.<\/p>\n\n\n\n
- \n
- Charge q on a body is always denoted by
\n$q = ne$
\nwhere n = any integer positive or negative and $e =1.602 \\times 10^{-19} C$ i. e., charge on an electron or proton.<\/li>\n - $Work\\; done = charge \\times potential $ or potential difference
\nMathematically,
\n$W=qV=q(V_2V_1)\\; Joule$<\/li>\n - $Electric\\; Current = \\frac{charge}{time}$
\nOr,
\n$I=\\frac{q}{t}\\;Ampere$<\/li>\n - Ohm\u2019s Law<\/strong>
\n$Resistance=\\frac{potential\\; difference}{current}$
\nOr,
\n$R=\\frac{V}{I}$<\/li>\n- Resistance in terms of resistivity<\/strong>
\n$R=\\frac{\\rho l}{A}$<\/li>\n- Resistance in series combination<\/strong>
\n$R_s=R_1+R_2+R_3+\u2026..$<\/li>\n- Resistance in parallel combination<\/strong>
\n$\\frac{1}{R_p}= \\frac{1}{R_1}+ \\frac{1}{R_2}+ \\frac{1}{R_3}+\u2026..$<\/li>\n- Electrical Power<\/strong>
\n$P=VI=I^2R=\\frac{V^2}{R}$<\/li>\n- Power in Series<\/strong>
\n$\\frac{1}{P_s}= \\frac{1}{P_1}+ \\frac{1}{P_2}+ \\frac{1}{P_3}+\u2026..$<\/li>\n- Power in parallel<\/strong>
\n$P_p=P_1+P_2+P_3+\u2026..$<\/li>\n- Electrical Energy<\/strong>
\n$E-Vit=I^2Rt=\\frac{V^2}{R}t$<\/li>\n- Heat produced<\/strong>
\n$H-Vit=I^2Rt=\\frac{V^2}{R}t$<\/li>\n<\/ol>\n\n\n\nSolved Examples <\/h2>\n\n\n\n
Question 1 <\/strong>
When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.
Solution
Here,
Battery Voltage $V= 12 V$
Current in circuit $I=2.5mA=2.5\\times 10^{-3}A$
Resistance of the resistor $R=?$ is to be calculated
We know that from Ohm\u2019s Law
$R=\\frac{V}{I}$
Putting in the values we get
$R=\\frac{12V}{2.5\\times 10^{-3}A}$
$=4.8\\times 10^{3}VA^{-1}=4.8\\times 10^{3}\\Omega $<\/p>\n\n\n\nQuestion 2<\/strong>
An electric heater of resistance $8\\Omega$ draws 15 A from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater.
Solution
It is given in the question that
Heater resistance $R=8\\Omega$
Current drawn, $I=15 A$
Time for which current is drawn
$t=2h=2\\times 60\\times 60 \\;sec=7200\\; sec$
Rate of heat developed $P=?$ which is to be calculated
We know that
Rate of heat developed = rate at which power is consumed.
$P=I^2R$
Putting in the value we get,
$P=(15 A)^2\\times 8\\omega=1800W$
The rate of heat developed is
$P=1800 J\/s$ Test Yourself on questions based on these formulas<\/p>\n\n\n\nTest Yourself <\/h2>\n\n\n\n
\n\n\n\nGeneral Instructions<\/strong><\/p>\n\n\n\n
- Your test contains multiple-choice questions with only one answer type of questions. There are a total of 5 questions<\/li>
- This is a 10 min test. Please make sure you complete it in the stipulated time<\/li>
- You can finish this test any time using the ‘View Results<\/em><\/strong>‘ button.<\/li><\/ul>\n\n\n\n
\n\n\n
- Resistance in terms of resistivity<\/strong>