JEE Physics Current Electricity Notes Set 02

Electric Current and Drift

Electric current is the rate of flow of charge through a cross-section, \( I = dq/dt \) (unit ampere). Charge passed in a time interval is the area under the I–t graph: \( q = \int I\,dt \). Conventional current is taken in the direction in which positive charge would move.
In a metal the free electrons move randomly at high speed and also drift slowly against the field. The drift velocity is \( v_d = eE\tau/m \), where \( \tau \) is the mean time between collisions. The current is
\[ I = neAv_d \]
where \( n \) is the number of free electrons per unit volume and \( A \) the area of cross-section. Current density is \( J = I/A = nev_d \). Drift speeds are only of the order of \( 10^{-4} \) m/s, yet the field travels through the wire almost instantly.
In steady flow the current is the same at every cross-section of a wire; the drift speed and the electric field vary inversely with the cross-sectional area.

Resistance, Resistivity and Ohm's Law

Ohm's law: for a conductor at constant temperature the current is proportional to the potential difference, \( V = IR \). A graph of V against I is a straight line through the origin for an ohmic conductor.
Resistance of a uniform conductor: \( R = \rho l/A \), where \( \rho \) is the resistivity of the material (unit ohm-metre). Conductivity is \( \sigma = 1/\rho \), so \( \rho\sigma = 1 \).
1. If a wire is stretched at constant volume, \( R \propto l^2 \) (a 20% increase in length raises R by 44%) and \( R \propto 1/r^4 \).
2. Bending a wire into a circle: two opposite points are separated by two parallel halves, giving \( R/4 \).
3. Cutting a wire into \( n \) equal parts and joining them in parallel gives \( R/n^2 \).
4. Resistance of a rectangular block depends on which pair of faces is used: \( R = \rho L/A \) with \( L \) the separation of the faces.

Temperature Dependence

For a metal, \( R_t = R_0(1 + \alpha t) \) with positive \( \alpha \); resistivity rises because the lattice vibrates more and collisions are more frequent. For semiconductors and insulators \( \alpha \) is negative because heating frees more charge carriers. Alloys such as manganin and constantan have very small \( \alpha \).
Two resistors in series have no net change with temperature if \( R_1\alpha_1 = R_2\alpha_2 \) in magnitude with opposite signs of the coefficients.

Combination of Resistors

1. Series: \( R = R_1 + R_2 + \dots \); the current is the same and voltages divide in proportion to resistance.
2. Parallel: \( 1/R = 1/R_1 + 1/R_2 + \dots \); the voltage is the same and currents divide inversely as resistance. The equivalent is less than the smallest branch.
3. n equal resistors r: series \( nr \), parallel \( r/n \).
4. Wheatstone bridge: balanced when \( P/Q = R/S \); no current then flows in the galvanometer and the bridge can be reduced by removing the middle arm.
5. Metre bridge: \( R/S = l/(100 - l) \) where \( l \) is the balance length from the end of the known resistance.

Kirchhoff's Laws

Junction rule: the sum of currents entering a junction equals the sum leaving it (conservation of charge).
Loop rule: around any closed loop the algebraic sum of potential changes is zero (conservation of energy). Going through a resistor in the direction of the current is a fall of \( IR \); going through a cell from negative to positive terminal is a rise of its e.m.f. Apply these two rules to get as many independent equations as unknown currents.

Cells, Heating Effect and Power

A cell of e.m.f. \( E \) and internal resistance \( r \) gives \( I = E/(R + r) \) and terminal voltage \( V = E - Ir \) when discharging (\( V = E + Ir \) when being charged). The terminal voltage equals the e.m.f. on open circuit; the short-circuit current is \( E/r \).
Cells in series: e.m.f. and internal resistance add. Identical cells in parallel: e.m.f. unchanged and internal resistance \( r/m \). For \( n \) cells in series in each of \( m \) rows the current is maximum when \( R = nr/m \).
Heating and power: \( H = I^2Rt = V^2t/R = VIt \) and \( P = VI = I^2R = V^2/R \). In series the larger resistance dissipates more power; in parallel the smaller resistance does. Power delivered to an external resistance is maximum when \( R = r \), equal to \( E^2/4r \). For a bulb rated \( P_0 \) at \( V_0 \), \( R = V_0^2/P_0 \) and at a different voltage \( P = (V/V_0)^2P_0 \). Energy is sold in units: 1 kWh = \( 3.6 \times 10^6 \) J.

Measuring Instruments and Potentiometer

1. Ammeter: galvanometer of resistance \( G \) with a small shunt \( S = I_gG/(I - I_g) \) in parallel; the ammeter resistance is lower than \( G \) and an ideal ammeter has zero resistance.
2. Voltmeter: galvanometer with a large series resistance \( R = V/I_g - G \); to extend the range from \( V \) to \( nV \) add \( (n-1)G \). An ideal voltmeter has infinite resistance.
3. Potentiometer: a long uniform wire carrying a steady current has potential drop proportional to length, \( V = xl \), with gradient \( x \) small for good sensitivity. It compares e.m.f.s, \( E_1/E_2 = l_1/l_2 \), and measures internal resistance \( r = R(l_1/l_2 - 1) \) where \( l_1 \) is the balance length on open circuit and \( l_2 \) with a shunt \( R \). At balance no current is drawn from the cell being tested.

Solved Examples

Example 1. A copper wire of cross-section 1 mm² carries 0.21 A. If the free-electron density is \( 8.4\times10^{28}\ \text{m}^{-3} \), find the drift speed.
Answer:
\( v_d = I/(neA) = 0.21/(8.4\times10^{28}\times1.6\times10^{-19}\times10^{-6}) = 1.56e-05\ \text{m/s} \).

Example 2. A wire of resistance 10 Ω is uniformly stretched to three times its length. Find its new resistance.
Answer:
Volume is constant, so \( R \propto l^2 \). \( R' = 9R = 90\ \Omega \).

Example 3. Find the equivalent resistance of a 3 Ω resistor in series with the parallel combination of 10 Ω and (3 Ω + 6 Ω).
Answer:
The branch 3 + 6 = 9 Ω is in parallel with 10 Ω: \( 10\times9/19 = 4.737\ \Omega \).
Total \( = 3 + 4.737 = 7.737\ \Omega \).

Example 4. A 12 V battery of negligible internal resistance is connected to a 6 Ω resistor in series with a parallel pair of 6 Ω and 3 Ω. Find the current.
Answer:
The parallel pair gives \( 6\times3/9 = 2\ \Omega \); total \( = 8\ \Omega \).
\( I = 12/8 = 1.5\ \text{A} \).

Example 5. A cell gives 0.5 A through 2 Ω and 0.25 A through 5 Ω. Find its e.m.f. and internal resistance.
Answer:
\( 0.5(2 + r) = 0.25(5 + r) \) gives \( r = 1\ \Omega \).
\( E = 0.5(2 + 1) = 1.5\ \text{V} \).

Example 6. A galvanometer of resistance 100 Ω has a full-scale current of 10 mA. Find the shunt to read up to 100 mA, and the series resistance to read up to 10 V.
Answer:
\( S = I_gG/(I - I_g) = 10\times100/90 = 11.11\ \Omega \).
\( R = V/I_g - G = 10/0.01 - 100 = 900\ \Omega \).