Tiêu đề 03. CURRENT ELECTRICITY.pdf ✅

Current density The current density at a point in a conductor is the ratio of the current at that point in the conductor to the area of cross–section of the conductor of that point. It is denoted by j i.e., j = A i i = Electric current A = Area of cross-section. Ohm’s law If there is no change in the physical state of a conductor (Such as temperature) then the ratio of the potential difference applied at its ends and the current flowing through it is constant i.e. I V V  I or V = I R; where, R is a constant. This is called 'Electrical resistance' of the conductor. This is true for metallic conductors only which have free electrons. The law is not applicable for ionized gases, transistors, semi– conductors etc. I V (a) I V (b) Diode I V (c) Semi conductor I V (d) Torch Bulb Units of resistance: ohm () 1 ohm = 1 volt / 1 ampere. Though electric current needs direction for its representation, yet it is scalar quantity. It is because, the current can be added algebraically. Only scalar quantities can be added algebraically not the vector quantities Note Did YOU KNOW? Electricity travels at the speed of light! That is more than 186000 miles per second! Q. How much current is present when 107 electrons per second run through a conducting wire? Sol. Flow of electrons, n t = 107 /second Therefore, current(I) = q t = ne t = n t × e = 107 × (1.6 × 10−19) = 1.6 × 10−12 A Area 'A' is normal to current 'I'. If A is not normal to I, but makes an angle q with the normal to current, then j = I Anormal = I A cos θ  I = j A cos q = . J =  2 × 106 A/m2  200 A/cm2 It is a VECTOR quantity Its direction is the direction of motion of the positive charges at that point. Units: ampere / meter2 (A/m2 ) Dimension: [M0 L –2T 0A] If, n = number of free electrons per unit volume of conductor. A = cross sectional area of conductor vd = Drift velocity. then I = neA vd and J = ne vd Note The speed of random motion of electrons is determined by temperature and is given by 1 2 mv2 = 3 2 kT v = ට 3kT m where m is mass of electron, T is absolute temp. and k is Boltzmann’s constant. (h) Electrons collide with the ions of metal while moving. The average time–interval between two successive collisions is called relaxation–time, denoted by t. The relations between relaxation time (t) and drift velocity (vd ) are given = – eEτ → m Note Drift velocity: An applied potential difference does not give an accelerated motion to electrons but simply gives them a small constant velocity (» 10–4 m/s) along the length of wire towards the end at higher potential. This is called Drift velocity of the electrons.

Combinations of Resistance (a) Series Combination R1 R2 R3 I V V1 V2 V3 Same current passes through each resistance. Voltage across each resistance is directly proportional to its value. V1 = IR1 , V2 = IR2 Sum of the voltages across resistances is equal to the voltage applied across the circuit i.e., V = V1 + V2 + V3 + .............. V = IR1 + IR2 + IR3 + .............. V I = R1 + R2 + R3 + .................. = R Where, R = equivalent resistance. For a series combination of two resistances R1 R2 V (A) equivalent resistance R = R1 + R2 (B) I = V / (R1 + R2 ) (C) V1 (voltage across R1 ) = IR1 = R1V R1+R2 (D) V2 (voltage across R2 ) = IR2 = R2V R1+R2 (b) Parallel combination: A B i1 i2 i3 R1 R2 R3 V There is same drop of potential across each resistance. Current in each resistance is inversely proportional to the value of resistance i.e. i1 = V R1 , i2 = V R2 , i3 = V R3 etc. Current flowing in the circuit is sum of the currents in individual resistances i.e. i = i1 + i2 + i3 , i = V R1 + V R2 + V R3 + ........  i V = 1 R = 1 R1 + 1 R2 + 1 R3 + .... where R = equivalent resistance. The equivalent resistance of parallel combination is lower than the value of lowest resistance in the combination. For a parallel combination of two resistances.... R1 R2 i1 i2 i V (i) i = i1 + i2 = V(R1+R2) R1R2 Comparative study of combination of resistor and capacitor- Sr.No Resistor Capacitor In series In parallel In series In parallel (i) (ii) (iii) (iv) (v) (vi) R1 R2 R3 V1 V2 V3 I + – E R = R1 + R2 + R3 V = V1 + V2 + V3 Current is same in all the resistances If n resistances, A B i1 i2 i3 R1 R2 R3 + – 1 R = 1 R1 + 1 R2 + 1 R3 V1 = V2 = V3 = V Currents through different but p.d. across C1 +q +q V1 C2 +q V2 C3 V3 1 C = 1 C1 + 1 C2 + 1 C3 V = V1 + V2 + V3 Charge is equal in all capacitors If n cap. is in series then C' = C/n C1 +q1 C –q1 2 C3 +q2 –q2 –q3 +q3 A B V C = C1 + C2 + C3 V1 = V2 = V Different capacitor diff. charge. Q. A wire 50 cm long and 1 mm2 in cross-section carries a current of 4 A when connected to a 2 V battery. Find the resistivity of the wire. Sol. Length (l) = 50 cm = 0.5 m; Area (A) = 1 mm2 = 1 × 10−6 m2 ; Current (I) = 4A and voltage (V) = 2 volts. Resistance(R) = V I = 2 4 = 0.5 Ω Resistivity(ρ) = R × A l = 0.5 × 1×10−6 0.5 = 1 × 10−6 Ω m