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# Currentelectricity

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prepwalk

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### Currentelectricity

1. 1. PHYSICS CURRENT ELECTRICITY Short revision series
2. 2. EMF  P.d between the terminals of the cell when it is not delivering any current in the external circuit  (i.e p.d across external and internal circuit)  E = IR + Ir  P.D is the work done when one coulomb of charge moves from one point to another.
3. 3. Resistivity and resistance ρ = AR/L R = ρL/A σ (conductivity) = 1/ρ = L/RA (siemens/m)  ρ = resistivity  A = Area  R = resistance
4. 4.  Eg find resistance of a wire of lenght 500cm and cross-sectional area of 2m2 , if the resistivity is 1 x 10-5 Ωm  R = ρL/A = 1x10-5 x 0.5 = 2.5 x 10-6 Ω 2  What is the conductivity? 1/ρ = 1/1 x10-5 = 1x105 Sm-1
5. 5. OHM’S law  The electric current flowing through a metallic conductor at constant temperature is directly proportional to the p.d between its ends  V α I V = Ki  V = IR *R is the constant  EMF = V +Ir = IR +Ir = I(R +r)  I = E/(R +r)
6. 6.  Eg find current in a voltmetre of 600 ohms resistor which reads 150V when connected to a battery of 20 0hms internal r.  V = IR I = V/R = 150/600 = ¼ A  What is the emf?  = I(R + r) = ¼ (600 + 20) = 155V
7. 7. Resistors in series and parallel  Resistors in series  R = R1 + R2 + R3 ....  Resistors in parallel  R = 1/R1 + 1/R2 + 1/R3 ....  2 resistances R1 and R2, the equivalent = R1R2 /R1 + R2  Eg you are given a circuit with 3 pieces of 5Ω resistors connected in series and 2 pieces of 2 Ω resistors connected in parallel. The resistance in the circuit =# R = 5+5+5+(2x2/2+2) = 16Ω
8. 8. Remember: balance conditions 1. A wheatstone bridge  R1/R2 = R3/R4 2. The metre bridge  X/R = l σ/(100 – l)σ  X and R are connected resistance  σ = resistance per unit lenght  l = lenght  Potentiometre  E1/E2 = l1/l2
9. 9. KIRCHOFF’S laws  For current  The algebraic sum of the currents flowing into any junction is zero. The sum of currents flowing into a junction = sum of currents flowing out  ΣI(in) = ΣI(out)  I1 + I2 + I3 = I4 + I5 I1 + I2 + I3 - I4 - I5 = 0 I1 I2 I3 I4 I5
10. 10.  For voltage  The algerbraic sum of the p.d in any closed loop = zero. The sum EMFs in any closed loop = the sum of voltage IR drops in the loop  ΣEMF – ΣIR = 0  ΣEMF = ΣIR
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