Exercise practice: Current Electricity
Current Electricity24 min readLesson 4 of 6
4.1 Drift speed and current
A copper wire of cross section 1.0 × 10^-6 m^2 carries a current of 1.0 A. If the free electron density is 8.5 × 10^28 m^-3, the drift speed follows from I = n e A v. Substitute to get v = 1 / (8.5 × 10^28 × 1.6 × 10^-19 × 1.0 × 10^-6), which is about 7.4 × 10^-5 m s^-1. Almost all of the current comes from the huge number of carriers, not from a fast drift.
The same wire is stretched to twice its length. The volume stays constant, so the area becomes half and the resistance becomes four times larger. The drift speed for the same current halves, because the area has halved. State the volume constraint before using the square law.
4.2 Resistance and resistivity
A wire of length 2 m and area 5 × 10^-7 m^2 has resistivity 1.6 × 10^-8 ohm metre. Its resistance is rho l / A = 1.6 × 10^-8 × 2 / (5 × 10^-7), which is 6.4 × 10^-2 ohm. Changing the current through it does not change the resistance as long as the temperature stays fixed.
If the temperature coefficient is 0.004 per degree and the temperature rises by 50 degrees, the resistance rises by about 20 percent, using R = R0 (1 + alpha Delta T) for a moderate temperature range. The linear formula is an approximation, so a question that gives large temperature changes is testing the direction of the change rather than the exact factor.
4.3 Cells, Kirchhoff and bridges
A cell of emf 12 V and internal resistance 1 ohm drives a current through a 5 ohm resistor. The current is 12 / 6 = 2 A, and the terminal voltage is 12 - 2 × 1 = 10 V. The lost volt is 2 V, which appears as heat inside the cell. To find the internal resistance from two measurements, use E = I1 (R1 + r) and E = I2 (R2 + r), set them equal and solve for r.
For a two-loop circuit, mark I1, I2 and I3 at the junction, write I3 = I1 + I2, then write the loop equations with the chosen sign convention. Solve the two linear equations together. Keep the units in volt and ohm, and check the solution by substituting back into both loops.
In a metre bridge, balance is found at 40 cm from the left end for an unknown resistance X in the right gap and a 6 ohm standard resistance in the left gap. The ratio X / 6 equals the ratio of the right length to the left length, that is 60 / 40, so X = 9 ohm. A potentiometer question gives a balance length proportional to emf; if a 2 V cell balances at 40 cm, then a length of 100 cm corresponds to 5 V.
Exam
In bridge and potentiometer problems, first write the balance condition, then substitute the lengths. In cell problems, write the terminal voltage expression and check whether the cell is charging or discharging before choosing the sign of I r.
4.4 Potentiometer and bridge numericals
A potentiometer wire of length 10 m carries a steady current, and a cell balances at 2.5 m. A second cell of unknown emf balances at 4 m. The ratio of emfs equals the ratio of balancing lengths, so E2 / E1 = 4 / 2.5 = 1.6. If E1 is 1.2 V, the unknown emf is 1.92 V. The comparison works because at balance no current is drawn from either cell, so each is measured at its true emf.
In a metre bridge the balance length is measured from one end, and the unknown resistance is R = S × (length on the other side) / (length on the same side as S). A balance at 60 cm with a 10 ohm standard in the left gap gives R = 10 × 40 / 60 = 6.67 ohm. Always state which gap holds the standard resistance, because reversing the gaps inverts the ratio.
4.5 Common mistakes in circuit exercises
Sign of the internal resistance changes when the cell is being charged rather than discharging. Kirchhoff loop equations need one consistent direction of traversal. In bridge problems, the balance condition uses the ratio of adjacent arms, not the ratio of opposite arms. Checking the units and the order of magnitude of the current catches most slips.
4.6 A two-loop Kirchhoff exercise
Two cells of 6 V and 4 V with internal resistances 1 ohm and 2 ohm are connected in a two-loop network with a 4 ohm common resistor. Label the current in the shared branch I1 and the two other branch currents I2 and I3. The junction rule gives I1 = I2 + I3. The left loop gives 6 = 1 I2 + 4 I1 and the right loop gives 4 = 2 I3 + 4 I1. Solve the three equations together. Substitution is faster than guessing, and checking each loop after solving catches sign errors.