JEE Physics Current Electricity Notes Set 06

Reducing Difficult Resistor Networks

Two resistors are in series only if exactly the same current must flow through both; they are in parallel only if the same two end points are shared. When a network is neither, use one of the methods below.
Equal potentials: points that must be at the same potential can be joined by a wire (or a resistor between them removed). Symmetry about the line joining the input and output terminals puts mirror-image points at equal potential.
Folding: if a circuit is a mirror image about a plane through the input and output, the plane can be treated as an equipotential and the circuit folded in half.
Current distribution: for a symmetrical network, assume the current splits equally at each symmetrical junction, label the currents, apply the loop law once, and use \( R_{eq}=\frac{V}{I} \).

Wheatstone Network and Its Balance Condition

In a four-arm bridge with arms P, Q (on one side) and R, S (on the other), the galvanometer between the two middle points reads zero when \( \frac PQ=\frac RS \). Then the middle arm can be removed or replaced by a wire without changing anything.
Cube and tetrahedron results with equal resistors R on every edge: body diagonal \( \frac{5R}{6} \); face diagonal \( \frac{3R}{4} \); edge \( \frac{7R}{12} \); any two corners of a tetrahedron \( \frac R2 \). These come directly from the symmetry rules above.
Capacitors behave as conductances: the same cube of capacitors C gives \( \frac{6C}{5} \) on the body diagonal, not \( \frac{5C}{6} \).

Infinite Ladder and Repeating Networks

For a ladder of identical sections that goes on forever, the resistance looking in from the left is the same as the resistance looking in after one section is removed. Call it x, write x in terms of itself (one series element, one parallel element) and solve the resulting quadratic, keeping the positive root.
Example: sections with series R and shunt R give \( x=R+\frac{Rx}{R+x} \), so \( x=\frac{1+\sqrt5}{2}R \).

Wires Under Stretching and Temperature Change

At constant volume, stretching a wire to n times its length makes its resistance \( n^2 \) times larger. A small fractional change in length changes R by twice that fraction; a small fractional change in radius changes R by four times that fraction with opposite sign.
For a metal, \( R_t=R_0(1+\alpha t) \). For two resistors in series, the effective temperature coefficient is \( \frac{R_1\alpha_1+R_2\alpha_2}{R_1+R_2} \); for equal resistances (series or parallel) it is nearly the mean \( \frac{\alpha_1+\alpha_2}{2} \).
If resistivity varies along a wire, slice it into thin pieces in series and integrate: \( R=\int\frac{\rho(x)\,dx}{A} \).

Power, Batteries and Grouping

A battery of emf E and internal resistance r delivers maximum power to a load R = r, where the power is \( \frac{E^2}{4r} \) and the efficiency is 50%. Efficiency \( \eta=\frac{R}{R+r} \) rises with R.
Grouping N identical cells as n parallel rows of m cells in series (N = mn): emf mE, internal resistance \( \frac{mr}{n} \). For maximum current through R, choose the grouping with \( \frac{mr}{n}=R \).
Cells in parallel with unequal emfs: \( E_{eq}=\frac{E_1r_2+E_2r_1}{r_1+r_2} \) lies between the two emfs and \( r_{eq} \) is smaller than either resistance. If one of the cells is ideal, its emf alone fixes the terminal voltage.
While a cell is being charged, the current flows in through its positive terminal and \( V=E+Ir \); the charging current is \( \frac{V_{charger}-E}{r} \).
For appliances in parallel, the powers add; the main fuse must carry the total power divided by the supply voltage. For bulbs in series the bulb of lower power rating has the higher resistance and may be overloaded.

Galvanometer, Ammeter and Voltmeter

A galvanometer of resistance G and full-scale current Ig becomes an ammeter of range I with a shunt \( S=\frac{I_gG}{I-I_g} \) in parallel (total resistance a little below S), and a voltmeter of range V with a series resistance \( R=\frac{V}{I_g}-G \).
Meters load the circuit. A finite-resistance voltmeter in parallel with a resistor R reads the voltage across \( \frac{RR_V}{R+R_V} \); an ammeter in series adds its own small resistance. The measured ratio V/I therefore differs slightly from the true R.
The full-scale current of a moving-coil galvanometer follows from \( NBAI=k\theta \) where k is the torsional constant.

Metre Bridge, Post Office Box and Potentiometer

Metre bridge: a known resistance R and an unknown X are in the two gaps; if the null point is at length l from the end next to R, \( X=R\frac{100-l}{l} \). Accuracy is best when l is between 40 cm and 60 cm; end corrections and interchanging R and X reduce errors.
Post office box: a compact Wheatstone bridge with ratio arms P and Q (10, 100, 1000 ohm) and a resistance arm R, so that \( X=R\frac{Q}{P} \). The battery key is pressed first and the galvanometer key afterwards, because pressing the battery key makes a transient self-induced current that would disturb the balance. With P = Q = 10 ohm, find the two values of R (1 ohm apart) between which the deflection reverses; then use P = 100 ohm to get one more decimal place.
Potentiometer: a long uniform wire carrying a steady current has potential gradient \( k=\frac{I\rho}{A} \). At balance no current flows in the secondary circuit, so a cell's emf is read directly as \( E=kl \). Comparison of emfs: \( \frac{E_1}{E_2}=\frac{l_1}{l_2} \). Internal resistance: \( r=R\left(\frac{l_1-l_2}{l_2}\right) \) where \( l_1 \) is the balance length on open circuit and \( l_2 \) with resistance R connected across the cell.
Sensitivity improves with a smaller gradient (longer wire or smaller driver current); it is more accurate than a voltmeter because it draws no current at balance.

Worked Examples

Worked Example 1. Find the resistance between opposite corners of a wire frame shaped like a cube with a resistor R on each edge.
Answer:
By symmetry the three edges leaving the first corner carry equal currents and their far ends are at one potential; the six edges between the middle layers are in two equal parallel groups of six; the three edges entering the last corner are another parallel group.
\( R_{eq}=\frac R3+\frac R6+\frac R3=\frac{5R}{6} \).

Worked Example 2. A potentiometer wire of length 100 cm and resistance 10 ohm is in series with a 2 V cell of negligible internal resistance and a resistance R. A 5 mV cell balances at 25 cm. Find R.
Answer:
Gradient: \( \frac{5\ \text{mV}}{25\ \text{cm}}=0.2\ \text{mV/cm} \), so the wire has 20 mV across it and carries \( \frac{20\ \text{mV}}{10\ \Omega}=2\ \text{mA} \).
\( R+10=\frac{2}{0.002}=1000\ \Omega \), so \( R=990\ \Omega \).

Worked Example 3. A cell of emf 2 V and internal resistance 0.5 ohm drives current through an external resistor R. For what R is the power in R equal to 1.5 W?
Answer:
\( P=\frac{E^2R}{(R+r)^2} \): \( 1.5(R+0.5)^2=4R \), so \( 3R^2-5R+0.75=0 \) after multiplying by 2.
\( R=\frac{5\pm\sqrt{25-9}}{6}=\frac{5\pm4}{6} \), giving \( R=1.5\ \Omega \) or \( R=\frac16\ \Omega \); the maximum power is 2 W at R = 0.5 ohm, so two values are possible.