Publicité

27 May 2020•0 j'aime## 0 j'aime

•146 vues## vues

Soyez le premier à aimer ceci

afficher plus

Nombre de vues

0

Sur Slideshare

0

À partir des intégrations

0

Nombre d'intégrations

0

Télécharger pour lire hors ligne

Signaler

Formation

This slide is part 3 of current electricity. We have uploaded lot more before. check that out and then study well. Thank you.

Ahamed Yoonus SSuivre

Teacher à Central university of tamilnaduPublicité

- SRV PUBLIC SCHOOL, Samayapuram ONLINE CLASS FOR GRADE 10 SUBJECT : PHYSICS TEACHER NAME :Mr. DANIELAMIRTHA RAJ
- 12. ELECTRICITY GRADE – X- SCIENCE - NCERT
- COMBINATION OF RESISTORS Symbol of resistor Series connection Parallel connection
- COMBINATION OF RESISTORS Resistors in series Resistors in parallel
- RESISTORS IN SERIES ‘I’ is constant ‘V’ varies
- RESISTORS IN SERIES When several resistors are joined in series, the resistance of the combination is always equal to the sum of their individual resistances Effective resistance is always greater than any individual resistance
- It is observed that the total current I, is equal to the sum of the separate currents through each branch of the combination. I=I1+I2+I3 Let RP be the equivalent resistance of the parallel combination of resistors.
- By applying Ohm’s law to the parallel combination of resistors. We have, I=V/Rp ; I1=V/R1 ; I2=V/R2 ; I3=V/R3 V/Rp= V/R1+ V/R2+ V/R3 1/Rp= 1/R1+ 1/R2+ 1/R3
- Thus we may conclude that the reciprocal of the equivalent resistance of a group of resistances joined in parallel is equal to the sum of the reciprocals of the individual resistances. Effective resistance is always lesser than any individual resistance when connected in parallel
- SERIES PARALLEL Current is Constant Voltage is Constant Voltage varies across different resistors Current flowing through the resistors varies with different resistors. Effective resistance will be greater than any individual resistors. RS = R1+ R2 + R3…….Rn Effective resistance will be less than any least individual resistors. 1 𝑅𝑝 = 1 𝑅1+1 𝑅2+1 𝑅3….1 𝑅𝑛
- picture
- ADVANTAGE OF PARALLEL ELECTRIC CIRCUITS Independent Components: • when you turn ON one gadget ,you don’t necessarily want to turn on all the others. • Individual switches can be provided to each device. Allows for additional components: • Parallel circuits allow components to be added in the circuit without changing voltage. • For example if you need an additional light in your home you can add one new bulb, it doesn't affect the current through other bulbs.
- FORMULA •If “n” resistors of equal resistances are connected in series then the effective resistance would be nR. •If “n” resistors of equal resistance are connected in parallel then the effective resistance would be R/n.
- PROBLEM 1 Q, If n equal resistors are connected in series and then connected in parallel, the ratio of the maximum to minimum resistance is__________ (a) n (b) 1 / n2 (c) n2 (d) 1 / n
- SOLUTION FOR PROBLEM 1 When n resistors are connected in series, Equivalent resistance Rn= nR (Maximum) When connected in parallel, Equivalent resistance Rp=R/n (minimum) The ratio of the maximum to minimum resistance is, nR /(R/n) = n2
- PROBLEM 2 Q, Three resistances each of 4 ohm are connected to form a triangle. The resistance between any two terminal is _____________
- SOLUTION Resistance between any two terminals = 4x8/4+8 = 32/12 =8/3 ohm 4𝛺 4𝛺 4𝛺 4𝛺 8𝛺
- PROBLEM 3 Q, Current through 3 ohm resistor is 0.8A then potential drop through 4 ohm resistor is______________ A, 9.6V B, 2.6V C, 4.8V D, 1.2V 6𝛺 3𝛺 4𝛺
- SOLUTION (C) Voltage across 3 ohm resistor = 3i1 = 3x 0.8 =2.4 V Voltage across 6 ohm resistor = 6i2=2.4V, i2=0.4 A 6𝛺 3𝛺 4𝛺 i i2 i1
- I = i1+i2 = 0.8+0.4 =1.2A Therefore voltage across 4 ohm resistor = 4i = 4x1.2 = 4.8 V
- HEATING EFFECT OF CURRENT Look this simple diagram A BULB K Bt
- If we switch ON the circuit , then the bulb will start to glow. In order to keep bulb glowing battery have to supply current continuously by expending its energy. Does all the energy spent by the source (battery) is utilized by the bulb? Yes or No?
- ANSWER IS “NO” • Because as the bulb started to glow, it consumes electrical energy and convert it into light energy but not all. • Some of the energy is dissipated as heat in and around the bulb.( you may have the experience while you touch an incandescent lamp which is kept ON for few minutes you feel it hot) • Another example : Electric fan becomes warm if it is used continuously for long time.
- APPLICATIONS OF THIS HEATING EFFECT • The energy wasted (in the form of heat) is an inevitable consequence of current. • Even though it is undesirable and unavoidable it gives insight that this heating effect can be utilized in some devices..(ex: Electric heater, electric iron etc..) • If the electric circuit is purely resistive then the entre energy from the source is converted in the form of HEAT.
- JOULES LAW OF HEATING Consider a current I flowing through a resistor of resistance R. Let the potential difference across it be V. V A K YX R I I + + + - --
- ELECTRICAL ENERGY ~ HEAT ENERGY • When electric current flows through resistors (which offers resistance) suffer obstruction. • Work has to be done to overcome this resistance , which is converted into heat energy. • If Q amount of charge flows through potential difference V, then • W= Q x V since [ W/Q = V]
- • W = Q x V ------------- (1) • I = Q / t • Q = I x t -------------- (2) From (1) & (2) W = V I t -------------- (3) By ohms law , V = I x R W = (IR) It = I 2 R t This work is converted into energy H = I 2 R t ---------------- (4) The above equation is called joules law of heating.
- PROBLEM - 4 •A current of 2A passing through conductor produces 80 J of heat in 10 seconds. The resistance of the conductor is (a) 0.5 Ω (b) 2 Ω (c) 4 Ω (d) 20 Ω
- SOLUTION OF PROBLEM 4 H = I2 R t 80 = (2)2 x R x 10 R = 80/40 = 2 ohm
- PROBLEM 5 •Two heater wires of same dimension are first connected in series and then in parallel to a source of power supply. What will be the ratio of heat produced in the two cases ?
- SOLUTION FOR PROBLEM 5 As heater wires are of the same dimensions, their resistance are equal, say R each. In series, their resultant resistance Rs=2R. Now, if connected to supply source of voltage V for time t, the heat produced, Hs= V2t/Rs = V2t/2R
- In parallel, the resultant resistance, Rp=R/2 And for same supply voltage V and for the same time t, the heat produced, Hp= V2t/Rp = 2V2t/R
- Hs= V2t/Rs = V2t/2R Hp= V2t/Rp = 2V2t/R Hs/Hp=1/4
- ELECTRIC FUSE
- ELECTRIC FUSE • It protects circuits and appliances by stopping the flow of any excessive high electric current. • It consists of a metal wire (ex: copper, aluminium….etc.) of low melting point. • It is placed always in series before the device. • If a current is larger than the specified value flows through the circuit , the temperature of the fuse wire increases then melts the fuse wire and breaks the circuit.
- • Fuse wire of various current capacities (1A,2A,5A…etc )is used for domestic purpose. • Example: • For an electric iron which consumes 1kW power when operates at 220V. In order to protect the device choose a suitable fuse. P = V x I or I = 𝑃 𝑉 = 1000𝑊 220 𝑉 = 4.5 A So 5A fuse must be used to protect the circuit.

Publicité