Chapter 3 - Current Electricity

Electric current is defined as rate of flow of charge: I = dQ/dt. SI unit Ampere A: 1 A = 1 C/s. Conventional current is direction of flow of positive charges from positive to negative terminal of battery. But actual electrons flow from negative to positive, opposite to conventional current. We still use conventional current direction for analysis.

Inside a conductor, free electrons move randomly with high speed ~10⁵ m/s at room temperature due to thermal energy. They collide with lattice ions frequently. Average time between collisions is called relaxation time τ ~10⁻¹⁴ s. Without electric field, average velocity of all electrons is zero, no net current.

When battery connected, electric field E is set up inside conductor. Field exerts force F=eE on each electron opposite to field (because electron negative). Electron accelerates but quickly collides and loses energy. Between collisions it has average drift velocity vd = a τ where a = eE/m is acceleration. So vd = eEτ/m. This drift is very small ~10⁻⁴ m/s, slow, but number of electrons huge so current significant.

Relation between current and drift: Consider wire length l cross-section A, number density n electrons per unit volume. Total electrons in volume Al = nAl, total charge Q=nAl e. Time taken for electrons to cross length l is t=l/vd. So current I=Q/t=nAl e / (l/vd)=n e A vd. So I=n e A vd. Current density J=I/A=n e vd vector in direction of E (for positive charges) or opposite for electrons but conventional along E. J=σE where σ conductivity. Mobility μ=vd/E=eτ/m drift velocity per unit field m²/Vs.

Example: Copper has n=8.5×10²⁸ m⁻³, wire area 1 mm²=1e-6 m², current 1 A, vd=I/(n e A)=1/(8.5e28×1.6e-19×1e-6)=7.4×10⁻⁵ m/s very small.

Ohm's law: At constant temperature, current through conductor is directly proportional to potential difference across it: V∝I or V=IR where R is resistance constant at given temperature. Resistance SI unit ohm Ω: 1 Ω = 1 V/A. Resistor opposes current.

Resistance depends on geometry and material: R=ρ l/A where l length, A cross-section area, ρ resistivity specific resistance Ωm material property independent of dimensions. Derivation: R=V/I=E l / (n e A vd) using E=vd/μ and vd relations gives ρ=m/(n e² τ). So ρ depends on n and τ.

Conductivity σ=1/ρ S/m siemens per meter, conductance G=1/R siemens.

Ohmic conductors obey Ohm's law V-I linear through origin slope=R. Examples metals at constant temperature. Non-ohmic V-I not linear examples diode, transistor, filament bulb, electrolyte, vacuum tube. For diode forward bias small voltage no current then rapid rise, reverse bias almost zero current.

Temperature dependence: Metals resistivity increases with temperature ρ=ρ0[1+α(T-T0)] α temperature coefficient positive ~0.004/°C for copper because τ decreases with temperature due to more lattice vibrations collisions. Semiconductors resistivity decreases with temperature α negative because n increases exponentially with temperature more free carriers even though τ decreases, effect of n dominates. Insulators also decrease. Alloys like manganin have very small α used for standard resistors stable.

Superconductors at very low temperature resistivity becomes zero below critical temperature Tc mercury 4.2K.

When current passes through resistor, electrons collide with lattice ions and give energy to lattice, lattice vibrates more temperature rises heat produced. This is Joule heating.

Work done by battery moving charge Q through potential difference V is W=QV. Power P=work per time=W/t=QV/t=VI since I=Q/t. So electrical power P=VI watt W. Using Ohm's law V=IR: P=I²R=V²/R. This power converted to heat in resistor.

Energy produced as heat in time t: W=P t=VI t=I²R t=V² t/R Joule J. This is Joule's law of heating. If resistor is pure resistor all electrical energy becomes heat. If circuit has motor, part becomes mechanical energy.

Rating of electrical appliances: Bulb rated 100W-220V means at 220V it consumes 100W power and resistance R=V²/P=220²/100=484 Ω. If two bulbs 100W and 60W same voltage 220V, 100W bulb has lower resistance and brighter at same voltage. In series connection, same current flows, P=I²R so higher resistance 60W bulb brighter in series. Important for exam.

Commercial unit of electrical energy is kilowatt-hour kWh: 1 kWh=1000W×3600s=3.6×10⁶ J. Electricity bill calculated in kWh units.

Resistors series: Connected end to end, same current I flows through each, total voltage V=V1+V2+...=I R1+I R2+...=I(R1+R2+...). So equivalent resistance Req=R1+R2+... sum larger than largest individual. Voltage divides proportionally to resistance V1=I R1 etc voltage divider.

Resistors parallel: Connected side by side between same two points, same voltage V across each, total current I=I1+I2+...=V/R1+V/R2+...=V(1/R1+1/R2+...). So 1/Req=1/R1+1/R2+... equivalent smaller than smallest. Current divides inversely to resistance I1=V/R1 etc current divider. Household wiring parallel so each appliance gets same voltage 220V.

Cells: Emf ε is work done per unit charge by cell in moving charge from negative to positive inside cell chemical energy to electrical. Unit Volt. Internal resistance r resistance of electrolyte and electrodes inside cell. Terminal voltage V is voltage across cell outside: When discharging (supplying current I) V=ε-Ir less than emf due to drop inside. When charging (current forced opposite) V=ε+Ir more than emf. When no current open circuit V=ε.

Combination cells series: n cells series εeq=ε1+ε2+... req=r1+r2+... if all identical ε r series n cells εeq=n ε req=n r current I=nε/(R+n r). Best when R>>n r external large.

Parallel cells: If identical cells parallel same emf εeq=ε req=r/n equivalent internal smaller current I=ε/(R+r/n). Best when external R small compared to internal. If cells different emf parallel complicated use Kirchhoff.

For complex circuits not simple series parallel, Kirchhoff's laws used.

First law - Junction rule: At any junction, sum of currents entering = sum of currents leaving. Σ I_in = Σ I_out. This is conservation of charge - charge cannot accumulate at junction, whatever comes must go out. If we take sign convention entering positive leaving negative sum zero Σ I=0.

Second law - Loop rule: Around any closed loop, algebraic sum of emfs = algebraic sum of IR drops. Σ ε = Σ I R. This is conservation of energy - work done by emfs in loop equals energy dissipated in resistors. Sign convention: Traversing loop, if go from - to + of battery emf +ε, + to - emf -ε. If traversing resistor along current direction IR drop -IR (potential decreases), opposite direction +IR (potential increases).

Example: Consider circuit with two batteries ε1 ε2 and resistors R1 R2. Apply junction at point: I1=I2+I3. Apply loop rule for two loops equations solve for currents.

Kirchhoff's laws allow solving any circuit with multiple batteries and resistors.

Wheatstone bridge is arrangement of four resistors P Q R S in diamond shape, battery connected across one diagonal (say BD), galvanometer across other diagonal (AC). Bridge is used to measure unknown resistance accurately.

Balance condition: Bridge is balanced when no current flows through galvanometer Ig=0. This happens when P/Q=R/S. Derivation using Kirchhoff laws: At balance Va=Vc so no potential difference between A and C galvanometer zero.

Proof: At balance Ig=0, current through P same as through Q (I1), current through R same as through S (I2). Potential drop P: I1 P and R: I1? Actually apply loops. Dividing gives P/Q=R/S.

Meter bridge is practical form of Wheatstone bridge. It has 1 meter uniform wire AC acting as two resistors P and Q proportional to lengths l1 and l2. Unknown resistance S and known R in other two gaps. Jockey moved along wire to find null point where galvanometer zero at length l1 from one end l2=100-l1 from other. Then R/S=l1/l2 so S=R l2/l1. Principle uniform wire resistance ∝ length R∝l. Accurate measurement of unknown resistance sensitive detects small changes at null point.

Potentiometer is long uniform wire AB length 4m with battery driving constant current through wire. Potential drops linearly along wire: Potential difference across length l is V=k l where k=potential gradient V/m constant if current constant and wire uniform.

Principle: Potential difference across a length of potentiometer wire is directly proportional to its length V∝l.

Applications:

1. Comparison of emfs: Two cells ε1 ε2 balanced at lengths l1 l2 from same end A when galvanometer shows zero no current through cell. At null ε1∝l1 ε2∝l2 so ε1/ε2=l1/l2. Compare emfs accurately.

2. Internal resistance of cell: Cell balanced without shunt at length l1 (ε∝l1). With shunt resistor R in parallel with cell, terminal voltage V balanced at l2 (V∝l2). Then ε/V=l1/l2. But V=ε R/(R+r) voltage divider. So r=R(l1-l2)/l2.

Advantages over voltmeter: Potentiometer draws no current at null point, so it measures true emf not terminal voltage, very accurate, independent of internal resistance at balance, no loading effect. Voltmeter draws some current and measures terminal voltage less than emf if internal resistance.

Potentiometer can also measure small potential differences and acts as variable resistor.