JEE Physics Current Electricity Notes Set 04

Current, Drift Velocity and Current Density

Electric current is the charge crossing a section per unit time, \( I=\frac{dq}{dt} \), with direction taken as that of positive charge flow. In a metal the free electrons drift opposite to the field with a tiny average speed \( v_d \) superposed on their random thermal motion. The current is \( I=neAv_d \), the current density is \( J=\frac IA=nev_d=\sigma E \), and \( v_d=\frac{eE\tau}{m} \) so that \( v_d\propto E \). Drift speed in copper is only of the order \( 10^{-4} \) m/s.
In a steady state the current is the same through every section of a wire, but the current density is larger where the wire is thinner, and so is the drift speed (\( v_d\propto 1/A \)).

Resistance, Resistivity and Temperature

Ohm's law: at constant temperature \( V=IR \). Resistance of a uniform conductor \( R=\frac{\rho l}{A} \), with resistivity \( \rho=\frac{m}{ne^2\tau} \). If a wire is stretched or compressed at constant volume, \( R\propto l^2 \) (equivalently \( R\propto\frac1{A^2}\propto\frac1{r^4} \)). A wire stretched to n times its length has n² times the resistance.
Temperature: \( R_t=R_0(1+\alpha\Delta T) \). Metals have a positive α, semiconductors a negative α (more carriers at higher temperature), and alloys such as manganin have very small α. For two resistors in series the effective coefficient is the resistance-weighted average \( \alpha=\frac{R_1\alpha_1+R_2\alpha_2}{R_1+R_2} \).
Resistor colour code: black 0, brown 1, red 2, orange 3, yellow 4, green 5, blue 6, violet 7, grey 8, white 9. The first two bands give digits, the third the multiplier power of ten, and the fourth tolerance (gold 5%, silver 10%, none 20%).

Combinations of Resistors

Series: same current, \( R=R_1+R_2+\dots \). Parallel: same voltage, \( \frac1R=\frac1{R_1}+\frac1{R_2}+\dots \). n equal resistors give nR in series and R/n in parallel; cutting a wire into n pieces and joining them in parallel gives R/n². A closed wire loop of resistance R, with the two points dividing it in the ratio a : b, has \( R_{eq}=\frac{R\,ab}{(a+b)^2} \).
Useful simplifications: points at the same potential can be joined or separated; no current flows in the arm of a balanced Wheatstone bridge; an infinite ladder is solved by assuming that adding one more section leaves the resistance unchanged.

Cells, EMF and Internal Resistance

The emf E of a cell is its terminal voltage on open circuit. With current I drawn, \( V=E-Ir \); while a cell is being charged, \( V=E+Ir \), so the terminal voltage then exceeds the emf. Series: \( E_{eq}=\sum E \), \( r_{eq}=\sum r \) (with signs for opposing cells). Identical cells in parallel: emf E, internal resistance r/n. Two unlike cells in parallel: \( E_{eq}=\frac{E_1/r_1+E_2/r_2}{1/r_1+1/r_2} \) and \( r_{eq}=\frac{r_1r_2}{r_1+r_2} \).
The power delivered to an external resistor R is \( P=\frac{E^2R}{(R+r)^2} \), a maximum of \( \frac{E^2}{4r} \) when \( R=r \) (maximum power transfer). With m rows of n identical cells, the current in R is greatest when \( R=\frac{nr}{m} \).

Kirchhoff's Laws

Junction rule: the algebraic sum of currents at a junction is zero (conservation of charge). Loop rule: the algebraic sum of potential changes around any closed loop is zero (conservation of energy). Going through a resistor along the current the potential falls by IR; going through a cell from the negative to the positive terminal the potential rises by E. Write one equation per independent junction and loop, then solve.

Electrical Power and Heating

Power dissipated: \( P=VI=I^2R=\frac{V^2}{R} \), and the heat in time t is \( H=I^2Rt \). Use \( I^2R \) for series elements (the larger resistor dissipates more) and \( V^2/R \) for parallel elements (the smaller resistor dissipates more). A bulb rated P watt at V volt has \( R=\frac{V^2}{P} \). Two bulbs in series on a supply smaller than the sum of their ratings share the voltage in proportion to their resistances.

Measuring Instruments

A galvanometer of resistance G and full-scale current \( I_g \) becomes an ammeter of range I with a shunt \( S=\frac{I_gG}{I-I_g} \) in parallel, and a voltmeter of range V with a series resistance \( R=\frac{V}{I_g}-G \). An ideal ammeter has zero resistance and is connected in series; an ideal voltmeter has infinite resistance and is connected in parallel. The figure of merit is the current per division, so full-scale current is the number of divisions times the figure of merit.

Wheatstone bridge: balanced when \( \frac PQ=\frac RS \); the galvanometer then reads zero and the cell and galvanometer may be interchanged without disturbing the balance. Meter bridge: \( \frac RS=\frac{l}{100-l} \) (with the unknown in the left gap); interchanging R and S removes end errors.
Potentiometer: the potential gradient is the wire's drop per unit length, \( k=\frac{IR_{wire}}{L} \). Comparing emfs: \( \frac{E_1}{E_2}=\frac{l_1}{l_2} \). Internal resistance of a cell shunted by R: \( r=R\left(\frac{l_1}{l_2}-1\right) \), where \( l_1 \) is the null length on open circuit and \( l_2 \) with R connected. A potentiometer draws no current at balance, so it measures emf accurately.

Worked Examples

Example 1. A 12 V battery of internal resistance 1 Ω is connected to resistors of 3 Ω and 6 Ω in parallel. Find the battery current and its terminal voltage.
Answer:
Parallel resistance: \( \frac{3\times6}{9}=2 \) Ω; total resistance \( 2+1=3 \) Ω.
\( I=\frac{12}{3}=4 \) A and \( V=E-Ir=12-4\times1=8 \) V.
The current through the 3 Ω and 6 Ω resistors is \( \frac83 \) A and \( \frac43 \) A.

Example 2. A wire of resistance 10 Ω is drawn out so that its length is doubled. Find the new resistance and, if the original is now cut into four equal parts joined in parallel, the resistance of that arrangement.
Answer:
Stretching at constant volume: \( R'=n^2R=4\times10=40 \) Ω.
Four equal parts of 2.5 Ω each in parallel: \( \frac{R}{n^2}=\frac{10}{16}=0.625 \) Ω.

Example 3. A galvanometer of resistance 40 Ω shows full-scale deflection for 5 mA. How can it be made into (a) a 0 to 2 A ammeter and (b) a 0 to 20 V voltmeter?
Answer:
(a) \( S=\frac{I_gG}{I-I_g}=\frac{0.005\times40}{1.995}=0.1 \) Ω in parallel.
(b) \( R=\frac{V}{I_g}-G=\frac{20}{0.005}-40=3960 \) Ω in series.