JEE Physics Current Electricity Notes Set 03

Electric Current and Drift Velocity

Current is the rate at which charge crosses a section of a conductor: \( I=\frac{dq}{dt} \). Its direction is taken as the direction in which positive charge would move.
Inside a metal the free electrons move randomly, and an applied field adds a small average velocity called the drift velocity \( v_d \). It is related to the current by \( I=neAv_d \), where n is the number density of free electrons and A is the cross-sectional area.
Drift speeds are very small (about 10⁻⁴ m/s in copper) even though a signal spreads through the wire almost instantly, because the field is set up along the whole wire at nearly the speed of light.
Current density: \( J=\frac{I}{A}=nev_d=\sigma E \). Relaxation time τ is the average time between two collisions and \( v_d=\frac{eE\tau}{m} \).

Ohm's Law, Resistance and Resistivity

For many conductors the current is proportional to the applied potential difference at constant temperature: \( V=IR \).
The resistance of a uniform conductor is \( R=\frac{\rho l}{A} \). In the free-electron picture \( \rho=\frac{m}{ne^2\tau} \) and conductivity \( \sigma=\frac1\rho \).
When a wire is stretched at constant volume, its length and area change together, so \( R\propto l^2 \) (equivalently \( R\propto \frac{1}{A^2} \)).
Colour code of carbon resistors: the first two bands give digits, the third gives the power of ten and the fourth gives tolerance (gold ±5%, silver ±10%, none ±20%).

Effect of Temperature

For metals, resistance rises roughly linearly with temperature: \( R_t=R_0(1+\alpha t) \), where α is the temperature coefficient of resistance. More lattice vibrations shorten the relaxation time, so resistivity increases.
For semiconductors the number of charge carriers rises quickly with temperature, so resistance falls and α is negative. Alloys such as manganin and constantan have very small α, so they are used for standard resistors.

Combination of Resistors

Series: the same current flows through every resistor and the voltages add. \( R_s=R_1+R_2+\dots \)
Parallel: the same voltage acts across every resistor and the currents add. \( \frac1{R_p}=\frac1{R_1}+\frac1{R_2}+\dots \); for two resistors \( R_p=\frac{R_1R_2}{R_1+R_2} \).
Useful tricks: a balanced Wheatstone bridge carries no current through its middle arm; points at the same potential can be joined or separated freely; an infinite ladder of identical sections is solved by assuming that removing one section leaves the resistance unchanged.

Cells, EMF and Internal Resistance

The emf E of a cell is its terminal voltage on open circuit. When a current I is drawn, the terminal voltage falls to \( V=E-Ir \), where r is the internal resistance. While a cell is being charged, \( V=E+Ir \).
Current through an external resistor R: \( I=\frac{E}{R+r} \).
n identical cells in series: emf nE, internal resistance nr. n identical cells in parallel: emf E, internal resistance r/n. A mixed grouping of m rows with n cells per row has emf nE and internal resistance \( \frac{nr}{m} \); the current through R is greatest when \( R=\frac{nr}{m} \).
Two cells in parallel: \( E_{eq}=\frac{E_1r_2+E_2r_1}{r_1+r_2} \) and \( r_{eq}=\frac{r_1r_2}{r_1+r_2} \).

Kirchhoff's Laws

Junction rule: the algebraic sum of currents at a junction is zero (charge is conserved).
Loop rule: the algebraic sum of potential changes around any closed loop is zero (energy is conserved). Going through a resistor along the current counts as −IR; going through a cell from its negative to its positive terminal counts as +E.

Electrical Energy and Power

Heat produced: \( H=I^2Rt=\frac{V^2}{R}t=VIt \) joule (divide by 4.2 for calories). Power: \( P=VI=I^2R=\frac{V^2}{R} \).
A bulb rated P watts at V volts has resistance \( \frac{V^2}{P} \) and draws \( \frac PV \) amperes. Bulbs in parallel at the rated voltage add their power.
Maximum power transfer: a source of internal resistance r delivers maximum power to an external resistance R = r, and then the efficiency is 50%. Two different loads R1 and R2 receive equal power when \( r^2=R_1R_2 \).

Galvanometer, Ammeter and Voltmeter

A moving-coil galvanometer of resistance G gives full-scale deflection for a small current Ig.
Ammeter: connect a small shunt S in parallel, \( S=\frac{I_gG}{I-I_g} \). Resistance of the ammeter is below S and G.
Voltmeter: connect a large resistance R in series, \( R=\frac{V}{I_g}-G \).
Current sensitivity is deflection per unit current; voltage sensitivity is deflection per unit voltage; their ratio gives G.

Meter Bridge and Potentiometer

Wheatstone bridge: balance when \( \frac PQ=\frac RS \). In the meter bridge, the balance point at length l from the left end gives \( \frac{P}{Q}=\frac{l}{100-l} \); interchanging P and Q changes l to 100 − l.
Potentiometer: a long uniform wire carries a steady current from a driver cell, so the potential drop is proportional to length. The potential gradient is \( k=\frac{V_{wire}}{L} \). Uses: compare emfs \( \frac{E_1}{E_2}=\frac{l_1}{l_2} \); find internal resistance \( r=R\left(\frac{l_1}{l_2}-1\right) \). It draws no current from the cell at balance, so it measures the true emf.

Worked Examples

Worked Example 1. Two cells of emf 6 V (r = 1 Ω) and 3 V (r = 2 Ω) are connected in parallel with like terminals together across a 4 Ω resistor. Find the current in the resistor.
Answer:
Equivalent emf: \( \frac{6\times2+3\times1}{1+2}=5\ \text{V} \); equivalent internal resistance: \( \frac{1\times2}{3}=\frac23\ \Omega \).
\( I=\frac{5}{4+2/3}=\frac{15}{14}\approx1.07\ \text{A} \)

Worked Example 2. A galvanometer of resistance 50 Ω gives full-scale deflection for 2 mA. How can it be converted into a 10 A ammeter?
Answer:
Shunt \( S=\frac{0.002\times50}{10-0.002}\approx0.01\ \Omega \), connected in parallel with the galvanometer.

Worked Example 3. A potentiometer wire of length 4 m and resistance 8 Ω is connected in series with a 2 V driver cell and a 12 Ω resistance (negligible cell resistance). Find the potential gradient.
Answer:
Current \( I=\frac{2}{8+12}=0.1\ \text{A} \); voltage across wire \( =0.1\times8=0.8\ \text{V} \).
Gradient \( =\frac{0.8}{4}=0.2\ \text{V/m} \).