CHAPTER: 3 CURRENT ELECTRICITY

CHAPTER: 3 CURRENT ELECTRICITY 1. Define electric current. Give its SI unit. *Current is the rate of flow of electric charge. I (t) = dq dt or I = q t SI unit is ampere (A), 1A = 1C 1s 2. Define current density (j). * It is the current flowing through unit area of cross section of a conductor. I J = S.I unit is A/m A 2 3. Define drift velocity (Vd). * In the absence of an external electric field, inside a conductor the free electrons are in random motion; and so their average velocity is zero. u1 + u2 +. un uavg = = 0 n When an electric field is applied the electrons move opposite to the electric field and get an average velocity. The average velocity (Vavg) acquired by an electron in the presence of an external electric field is called drift velocity 4. Define relaxation time (τ). * Relaxation time is the average time (tavg) interval between two successive collisions. 5. Derive a relation between drift velocity (Vd) and relaxation time (τ) When an electric field is applied the electrons move opposite to the electric field and force on the electron is, F= (- e )E F= qe Where, -e is the charge of electron and m- mass of electron. Acceleration of electron, a = F m = - ee m We have, v= u + at Vavg = uavg + a tavg Vd = 0+ a τ Vd = - ee m τ ---------(1) 6. Define mobility * Vd α E Vd = μ E Mobility ( μ ) is defined as the ratio of magnitude of the drift velocity to electric field strength. μ = Vd E = - e τ m --------(2) 7. Derive a relation between drift velocity and current. * Consider a conductor of length and number density of electrons n. [n = number of electrons per volume] Volume of the conductor = A No.of electron = n x volume = n A Total charge, Q = n A e I= Q t I = n A e ----(3) t = Vd t i.e., I = n A Vd e Current density = j = I A j = n Vd e ------(4) Where, n- number density of electrons. 1 P a g e

8. State Ohm s law. * Ohm s law states that at constant temperature the current flowing through a conductor is directly proportional to potential difference between the ends of the conductor. Potential difference α Current Vα I V= I R where, R Resistance of the conductor. 9. What is meant by resistance? * It is the ability of a conductor to oppose electric current. SI unit of resistance is ohm. 10. Define conductance (C). *Conductance is the reciprocal of resistance. C = 1 R SI unit = 1 Ω = 1 Ω -1 = ohm -1 = mho. 11. What are the factors which affect the resistance of a conductor? 1. Nature of the material of the conductor. 2. Length of the conductor (directly proportional) 3. Area of cross-section A (Inversely proportional) 4. Temperature (directly proportional) i.e. R α A R= ρ A Where, ρ is the resistivity of the material of the conductor. S.I. unit of Resistivity ohm-metre (Ω - m) 12. Define resistivity of the material of a conductor. Resistivity, ρ = RA Put A = 1 m 2, = 1m ρ = RA = Rx 1 = R 1 Resistivity of the material of a conductor is defined as the resistance of the conductor having unit length and unit area of cross section. 13. Define conductivity (σ) Conductivity is the reciprocal of resistivity. σ = 1 ρ SI unit of conductivity is Ω -1 m -1 or mho m -1 ( m -1 ) 14. What are the factors which affect resistivity? *(i). Nature of the material of the conductor. (ii). Temperature. 15. Derive the relation for resistivity in terms of relaxation time and hence deduce ohm s law. We have, I = n A Vd e -----(1) eq. (2) in (1), and Vd = ee m τ ----------(2) I = n A e [ ee m τ ] I= n A e2 E τ m But, V = E x V= E x d I= n A e2 V τ m R= V I = m n A e 2 τ ---------(3) R= ρ A = m n A e 2 τ Resistivity, ρ = m n e 2 τ Conductivity, σ = 1 ρ σ = n e2 τ m 2 P a g e

Deduction of ohm s law:- From eq. (3), R= V I = m n A e 2 τ V m l i.e, = I n e 2 τ A V I = a constant. This is ohm s law 16. Why copper is used as for making connecting wires? * Copper has low resistivity. 17. Why Nichrome is used as heating element of electrical devices? * Nichrome has i) High resistivity ii) High melting point. 18. Why aluminium wires are preferred for overhead power cables? * Aluminium has low resistivity. It is cheaper and lighter. 19. What are Ohmic and Non-ohmic substances. Ohmic substances They are the substances which obey ohms law. For these substances V-I graph is linear. Eg: Metals Non ohmic substances They are the substances which do not obey ohm s law. For these substances V-I graph is nonlinear. Eg:-Semiconductors, diodes, Vacuum tubes, electrolytes etc. 20. Ohm s law is not a universal law. Explain. * All materials do not obey Ohm s law. Metals obey ohms law while semiconductors, electrolytes, diodes etc. do not obey Ohm s law. So Ohm s law is not a universal law. 21. Explain how resistance depends on temperature. When temperature increases, resistances of materials change. R 2 = R1 [1 + α (T2 T1) ] -------(1) R1 resistance at the reference temperature T1. R2 resistance at temperature T2. Temperature coefficient of resistance. α = R 2 - R1 R1 (T2 - T1 ) Unit of is 0 C -1 or K -1 *(i).the value of is positive for metals resistance increases with increase of temperature. (ii).the value of negative for semiconductors resistance & resistivity decreases with increase of temperature & (iii). Nearly zero for insulators. Note:- ρ T = ρ0 (1+ t) 3.5. (page: 127). At room temperature (27.0 C) the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of the resistor is 1.70 10 4 C 1. R 2 - R1 α = R1 (T2 - T1 ) T2 T1 = R 2 - R1 R1 3 P a g e

T2 27 = 117-100 100 (1.70 10 4) T2 27 = 1000 T2 = 1027 0 C At 1027 0 C, the reisstance of the element is 117 Ω. Superconductivity When temperature decreases resistance of the material also decreases. But for some material when the temperature reaches a certain critical value (Tc) the resistance of the material completely disappears. The material begins to behave as a SUPER CONDUCTOR which does not offer any resistance to the flow of electrons. The critical temperature for mercury is 4.2K and for Lead is 7.25K. 22. Why materials like constantan and manganin are used to make standard resistances? * The temperature coefficient of resistance of these materials is nearly zero. Therefore, their resistances do not change considerably with temperature. Moreover these materials have high resistivity. 23. Give the codes for different coloured rings of carbon resistors. 24. Determine the resistance of the carbon resistor in the following figure. Resistance R = (47 10 2 ± 5% )Ω *Colour coding of a carbon resistance is used to find the value of resistance. 24. Derive expressions for the effective resistance when three resistors are connected in (i) series (ii) parallel. *(i).resistors in series: Consider resistances R1, R2 and R3 connected in series with a voltage V. 4 P a g e

In series circuit the current is the same, but the voltages across different resistors are different. The applied voltage, V = V1 + V2 + V3 ---- (1) But V = I Rs VI = I R1, V2 = I R2 and V3 = I Rs eq.(1) I Rs = I R1 + I R2 + I R3 IRs = I (R1+R2+R3) Rs = R1 + R2+ R3 If there are n resistors. Rs = R1 + R2 + R3 + + Rn. *(ii).resistors in parallel: Ans: In the external circuit, the current flows from the positive terminal of the cell to the negative terminal. But inside the cell the current flows from ve terminal to the +ve terminal. Internal resistance of a cell is the resistance offered by the electrolyte and electrodes of the cell. (Or) Resistance offered by the cell when an electric current flows through it. Relation between internal resistance (r), emf (E) and terminal potential difference (V). When a cell is connected to a resistor R, the current flows through R from the positive terminal of the cell to the negative terminal. Through the cell the current flows from negative terminal to the positive terminal. In parallel circuit voltage across each resistor is the same but the currents are different. I = I1 + I2 + I3. (2) But I = V Rp I1 = V R1, & I2 = V R2 eqn (2) V = V + V + V Rp R1 R1 R3, I3 = V R3 1 = 1 + 1 + 1 Rp R1 R2 R3 *In the circuit shown, R 1 = 4Ω, R 2 = R 3 = 15Ω, R 4 = 30Ω and E = 10V. Calculate the equivalent resistance of the circuit and the current in each resistor. (CBSE DELIHI 2011) 25. Define internal resistance (r) of a cell. Effective resistance = R + r Where r Internal resistance of the cell E Current I = R+r I(R + r) = E I R + I r = E But, IR = V, V Terminal Voltage (across R) V + Ir = E --------(1) Ir = E V r = E V I If I = 0 then from eqn(1) V= E (no current is drawn from the cell) 5 P a g e

Emf (E) of a cell:- The terminal potential difference is equal to the emf of the cell when no current is drawn from the cell. 26. What are the factors which affect the internal resistance of a cell? * (i) The nature of the electrolyte and the electrodes. (ii) The distance between the electrodes. (iii) Temperature (iv). Area of the electrode etc., 27. Derive expressions for the current due to a combination of cells. *i) Series:- (i). The equivalent emf of a series combination of n cells is just the sum of their individual emf s, and (ii). The equivalent internal resistance of a series combination of n cells is just the sum of their internal resistances. Consider n cells connected in series with a resistor R. Effective resistance = R + nr Current I = emf R = ne R + nr ii) Parallel:- Consider cells connected in parallel with an external resistance. VAB = VA VB = E1 I r1 -----------(1) VBC = VB VC = E2 I r2 ------------(2) eqn. (1) + (2) VA VC = VAC = E1 I r1 + E2 I r2 VAC = E1 + E2 I (r1 + r2) ---- (3) If Eeq is the resultant emf between A and C and req is the effective resistance then eqn (3) Veq = Eeq I req. Where Eeq = E1 + E2 & req = r1 + r2 If we have n number of cells of equal emf E then, Total emf Eeq = E +E +. + E = ne Total Internal resistance req = r + r + r + + r = nr Along ACB, VAC = V = E1 I1 r1 I1 = E 1 - V -----------( 1 ) r1 Along ADB, VAD = V = E2 I2 r2 6 P a g e

At A, I2 = E 2 - V r2 I = I1 + I2 I = E 1 - V r1 I = E1 r1 I = [ + E2 r2 E1 r2 + E2r1 r1+ r2 V [ ] = [ -----------( 2 ) + E 2 - V r2 - v [ 1 r1 + 1 r2 ] r1+ r2 ] - v [ ] E1 r2 + E2r1 Multiplying both sides by, E1 r2 + E2r1 ] I r1+ r2 V = - I ( r1 + r2 r1+ ) -----(3) r2 This is in the form V= Veq = Eeq I req.--------(4) Where, Eeq = req = E1 r2 + E2r1 r1 + r2 r1+ r2 If n cells of emf E connected in parallel with an external resistance, ; 28. State Kirchhoff s rules. *1st rule - (Current rule or Junction rule) Kirchhoff s first rule states that In a closed circuit the algebraic sum of the currents meeting at a junction is zero In other words, The total current entering a junction is equal to the total current leaving the junction. I1 + I2 + I3 I4 I5 = 0 Or I1 + I2 + I3 = I4 + I5 Kirchhoff s 1 st rule is a statement of law of conservation of charge. 2nd rule [Voltage Rule or Loop rule or mesh rule] Kirchhoff s second rule states that In a closed circuit the algebraic sum of the voltage drop is zero. E r + Er + Er Total emf Eeq = = E r + r+ r Total Internal resistance 1 = 1 + 1 req r r +.+ 1 r = n r r req = n Effective resistance = R + r n = nr + r n In the closed circuit ABFGA - I2R2 - I1R1+ E =0 Similarly for the mesh [closed circuit] BCDFB, -I3R3+ I2R2 = 0 7 P a g e

Kirchhoff s second rule is a statement of law of conservation of energy. Wheatstone s bridge 29. What is the use of a Wheatstone s bridge? * It is used to find the resistance of a given wire. 30. Derive the balancing condition of Wheatstone s bridge. Explain how the unknown resistance can be determined. Circuit Details Consider 4 resistances P, Q, R and S connected in the form of a bridge. A cell of emf E is connected between the terminals A and C. A galvanometer is connected between the terminals B and D. Balancing Condition Applying Kirchhoff s 2nd rule in the loop ABDA, - I1P - IgG + I2R = 0. (1) Again in the mesh BCDB, - I3Q + I4S + IgG = 0. (2) The resistance S is adjusted so that current through the galvanometer is made zero. ie., Ig = 0 Then, I1 = I3 and I2 = I4 eq.(1) - I1P + I2R = 0; I1P = I2R (3) Eq.(2) - I3Q + I4S = 0; I3Q = I4S.. (4) Eqn. (3) I1P (4) I3Q = I2R I4S But when the bridge is balanced, I1=I3 and I2= I4, P Q = R S This is Wheatstone s principle. This is the balancing condition of Wheatstone s bridge. Procedure The wire whose resistance is to be determined (Q) is connected across B and C. The value of resistance S is adjusted so that the current through the galvanometer is zero. Now the bridge is balanced and we have P Q = R S Since the values of P, Q, R and S are known, we can calculate the value of Q using the equation Q = P S R Meter Bridge 31. What is the use of a meter bridge? * It is used to find the resistance of a given wire. 32. Using a neat circuit diagram explain the circuit details, principle of a meter bridge. And give the procedure to find an unknown resistance. Circuit Details Meter Bridge consist of a one meter long resistance wire (made of constantan or manganin) fixed on a 8 P a g e

wooden board. A cell of emf E and a key are connected between the terminals A and B. A jockey is connected to the terminal C through a galvanometer. The unknown resistance is connected in the left gap and a resistance box is connected in the right gap. Principle It works on the Wheatstone s bridge principle. At the balancing condition, P Q = R S No current flows through the galvanometer. X R = l r (100- l)r Where r- resistance per unit length. X R = l (100- l) Procedure The key is closed. A suitable resistance (say 1Ω ) is taken from the resistance box. The jockey is moved from A towards B, until the galvanometer shows the zero deflection. The balancing length (AJ = l ) is measured. Unknown resistance X is determined using the equation, X = X = R l (100- l) R l (100- l) X is determined for different values of R (say 2Ω, 3Ω, -----). Now the experiment is repeated after interchanging X and R. Resistivity Resistivity, ρ = R A, l Here, R = X, A = πr 2 l is the length and r is the radius of the given wire whose resistance is to be determined. Potentiometer 33. What are the uses of a potentiometer? *Potentiometer is used (i) To compare the emf of two cells (ii) To find the internal resistance of a cell Consider a resistance wire of uniform area of cross section carrying a current I, The potential difference across length l of the wire is given by, V = IR V = I l r, Where r is the resistance per unit length of the wire V = (Ir) l V = k l V α l, Where k = Ir Principle When a steady current (I) flows through a wire of uniform area of cross section, the potential difference between any two points of the wire is directly proportional to the length of the wire between the two points. 34. Using a neat connection diagram, explain how potentiometer can be used to compare the emf of two cells. Circuit details The primary circuit consists of potentiometer, cell of emf ε, key K and a rheostat. The secondary circuit consists of two cells ε1 and ε2, two-way key, galvanometer, high resistance and Jockey. 9 P a g e

Connection diagram Connection Diagram The key K in the primary circuit is closed. The cell E1 is brought into the circuit by putting the key. Jockey is moved from A towards B until the galvanometer shows zero deflection. The balancing length AJ= l 1 is found out. Now by the principle of potentiometer, E1 α l 1 -------(1) The cell E1 is replaced by the cell E2 by putting the key. Again balancing length is found out. E2 α -------(2) Eqn. (1) E1 E2 = l 1 (2) If one of the cells is a standard cell ( say E1), we can determine the emf of the other, E2 = E1 l 1 35. Using a neat connection diagram explain how potentiometer can be used to find the internal resistance of a cell. * Circuit details The primary circuit consists of a potentiometer, cell of emf E, key K and a rheostat. The secondary circuit consist of a cell of emf E, a galvanometer, Jockey, a resistance box R and key K. Procedure The key K in primary circuit is closed. The balancing length l 1 is found out. E α l 1 (1) Now the key K in the secondary is closed. The balancing length found out. Then we can write, V α E R R +r α -----------(2) I = V R = is E R +r eqn (1) (2) = V= I R = E R R +r E ER R+r r = = l 1 R +r R = l 1 R + r = R ( l 1 r = R l 1 r = R l 1 - R R (l 1 - ) ) - R Using this equation we can calculate the internal resistance of the given cell. 10 P a g e

36. Why potentiometer is preferred over voltmeter for measuring emf of a cell? Ans: If we measure emf of a cell using a voltmeter, a small current is drawn by the voltmeter. So we will not get the actual emf of the cell. But if we measure the emf using a potentiometer, while measuring emf, since null deflection method is employed, no current flows from the cell. So we will get the actual emf of the cell. 37. State Joule s law of heating. * Joule s Law states that the heat produced in a current carrying conductor is proportional to the square of the intensity of electric current, the resistance of the conductor and the time for which the current flows. H = I 2 Rt S.I. unit of heat is joule (J) H= I 2 Rt H= VIt H= V2 t R H= P t 38.Define electric power. * Electric power is defined as the electric energy consumed per second P= I 2 R P= VI H= V2 R SI unit of electric power is watt (W) 39. Which is the commercial unit of electric energy? What is its relation with joule. * The commercial unit of electric energy is kilo watt hour. 1kWh=1000W x 3600s =3.6 x 10 6 J 11 P a g e