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Electricity Part 2

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Published in: Physics
1,271 Views

For class 10

Anita Y / Gurgaon

7 years of teaching experience

Qualification: Masters in physics

Teaches: Mathematics, Physics

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  1. Electricity part 2 - ohm's law and resistance By Anita Yadav M.sc Physics Ohm's Law Ohm's Law states that the potential difference between two points is directly proportional to the -R electric current. This means; current. or, Val ohm's law formula potential difference V varies as electric 1 1 (ii) RI — —a — (iii) Where R is constant for the given conductor at a given temperature and called resistance. Resistance is the property of through it. Sl Unit of conductor which resists the flow of electric current This means 1 = IA resistance is ohm. Ohm is denoted by Greek letter 'Q'. 1 ohm (Q) of Resistance (R) is equal to the flow of 1 A of current through a conductor between two points having potential difference equal to 1 V. one ohm From the expression of Ohm's Law it is obvious that electric current through a resistor is inversely proportional to resistance. This means electric current will decrease with increase in resistance and vice versa. The graph of V (potential difference) versus I (electric current) is always a straight line.
  2. Electric ctnrent (0 Graph of Potettid Difference (V) Vs Electric Currert (l) Resistance Resistance is a property of conductor due to which it resists the flow of electric current through it. Component that is used to resist the flow of electric current in a circuit is called resistor. In practical applications, resistors are used to increase or decrease the electric current. Variable Resistance: The component of an electric circuit which is used to regulate the current; without changing the voltage from the source; is called variable resistance. Rheostat: This is a device which is used in a circuit to provide variable resistance. Cause of Resistance in a Conductor: Flow of electrons in a conductor is electric current. The particles of conductor create hindrance to flow of electrons; because of attraction between them. This hindrance is the cause of resistance in the flow of electricity. Resistance in a conductor depends on nature, length and area of cross section of the conductor. Nature of material: Some materials create least hindrance and hence are called good conductors. Silver is the best conductor of electricity. While some other materials create more hindrance in the flow of electric current, i.e. flow of electrons through them.
  3. Such materials are called bad conductors. Bad conductors are also known as insulators. Hard plastic is the one of the best insulators of electricity. Length of conductor: Resistance R is directly proportional to the length of the conductor. This means, Resistance increases with increase in length of the conductor. This is the cause that long electric wires I —(iii) create more resistance to the electric current. Thus, Resistance (R) oclength of conductor (l) or R OCI --------(i) or, R oc From equation (i) and (ii) 11 Area of cross section: Resistance R is inversely proportional to the area of cross section ( A) of the conductor. This means R will decrease with increase in the area of conductor and vice versa. More area of conductor facilitates the flow of electric current through more area and thus decreases the resistance. This is the cause that thick copper wire creates less resistance to the electric current. Thus, resistance oc1/Area of cross section of conductor (A) Where p (rho) is the proportionality constant. It is called the electrical resistivity of the material of conductors.
  4. The Sl unit of resistivity: Since, the Sl unit of R and Sl unit of length is m. Hence Thus, Sl unit of resistivity (p) is Q m From equation (iii) RA n—, pl — p + — — — (iv) Diameter = 0.6 mm is Q, Sl unit of Area is m2 unit of resistivity (p) f) x m2 Materials having resistivity in the range of 10-8 Q m to 10-6 Q m are considered as very good conductors. Silver has resistivity equal to 1.60 X 10-8 Q m and copper has resistivity equal to 1.62 X 10-8 Q m. Rubber and glass are very good insulators. They have resistivity in the order of 1012 Q m to 1017 Q m. Resistivity of materials varies with temperature. Example 1: What will be the resistivity of a metal wire of 2 m length and 0.6 mm in diameter, if the resistance of the wire is 50 Q. Solution: Given, Resistance ( R) = 50 Q, Length ( I ) = 2 m
  5. Hence, radius = 0.3 mm Resistivity (p) = ? = 3 x 10-4 m Now, area of cross section of wire = Tt r2 or, A = 3.14 x (3 x 10-4)2 or, A = 28.26 x 10-8 rn2 = 2.826 x 10-9 rn2 Resistance: Part 2 Resistance Of A System of Resistors: Resistors are joined in two ways, i.e. in series and in parallel. Resistors in Series: When resistors are joined from end to end, it is called in series. In this case, the total resistance of the system is equal to the sum of the resistance of all the resistors in the system. Let total resistance = R Resistance of resistors are RI, R2, R3
  6. Therefore, R - Resistors in parallel: When resistors are joined in parallel, the reciprocal of total resistance of the system is equal to the sum of reciprocal of the resistance of resistors. R. resistance in series Fig: Resistance in Series Fig: Let total resistance = R Resistance of resistors are RI, R2, R3 Resistance in Parallel
  7. Example 1: There are three resistors joined in series in a system having resistance equal to 10 Q, 20 Q and 30 Q respectively. If the potential difference of the circuit is 240 V, find the total resistance and current through the circuit. Solution: Given, RI - Total resistance (R) =? Current through the circuit ( I ) =? According to Ohm's Law Total resistance in series (R) = Sum of resistance of all resistors Or, R = 10 Q We know that electric current I = V/R Thus, total resistance (R)= 60 Q Current through the circuit = 4 A Example 2: There are two electric lamps M and N which are joined in a series having resistance equal to 15 Q and 20 Q respectively. If the potential difference between two terminals of electric circuit is 220V, 1 1 1 1 sum total resistance in parallel - 10 Q, R2 = 20 Q, R3 = 30 Q and V = 240 V 1
  8. find the total resistance and electric current through the circuit. Also find the potential difference across the two lamps separately. Solution: Given, resistance (RI) of one electric lamp, M Resistance (R2) of other electric lamp, N = 20 Q Potential difference (V) through the circuit = • 220 v Electric current (l) through the circuit =? = 15.2 Q Potential difference through each of the electric lamp =? According to Ohm's Law; total resistance in series = Sum of resistance of all resistors = 15.2 Q + 20 Q = 35.2 Q Electric Current I = v/R = 220 V + 35.2 Q = 6.25 A Potential difference (VI) across electric lamp M Potential difference (V2) across electric lamp N Thus, electric current through the circuit = = 15.2 Q + 6.25 A = 2.432 V = 6.25 A = 3.2 V • 6.25 A Potential difference through electric lamp M = 2.432 v
  9. Potential difference through electric lamp N = 3.2 V

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