Current Electricity

Electric current is the rate of flow of charge: I = Q/t, where Q is charge (C) and t is time (s); [I] = [A], SI unit: ampere (A). At the microscopic level, I = neAvd, where n is the number density of free electrons ([n] = [$L^{-3}$], SI unit: /$m^{3}$), e is electron charge, A is the cross-sectional area, and vd is the drift velocity.

Drift velocity is derived from the force on free electrons in an applied electric field: vd = eE $\tau$/m, where E is the electric field, $\tau$ is the mean relaxation time ([$\tau$] = [T], SI unit: s), and m is the electron mass.
Numerically, vd is extremely small (~$10^{-4}$ m/s), yet current appears instantaneous because the electric field propagates at nearly the speed of light; [vd] = [L $T^{-1}$], SI unit: m/s.
Current density J = I/A = nevd = $\sigma$ E; [J] = [A $L^{-2}$], SI unit: A/$m^{2}$.

Ohm's law states V = IR, where V is potential difference (volt), I is current (ampere), and R is resistance; [R] = [M $L^{2} \; T^{-3} \; A^{-2}$], SI unit: ohm.
Resistance depends on geometry: R = $\rho$ l/A, where $\rho$ is resistivity ([$\rho$] = [M $L^{3} \; T^{-3} \; A^{-2}$], SI unit: ohm m) and $\sigma$ = 1/$\rho$ is conductivity ([$\sigma$] = [$M^{-1} \; L^{-3} \; T^{3} \; A^{2}$], SI unit: S/m).
Temperature dependence: R = $R_{0}$(1 + $\alpha$ DeltaT), where $\alpha$ is the temperature coefficient; for metals $\alpha$ > 0 (resistance increases with temperature), for semiconductors $\alpha$ < 0 (resistance decreases).

Power dissipation: P = VI = $I^{2}$ R = $V^{2}$/R; [P] = [M $L^{2} \; T^{-3}$], SI unit: watt (W). Use $I^{2}$ R when current is known (series circuits), $V^{2}$/R when voltage is known (parallel circuits).

For resistors: Series $R_{eq}$ = R₁ + R₂ + ... (same current through all, voltage divides). Parallel 1/$R_{eq}$ = 1/R₁ + 1/R₂ + ... (same voltage across all, current divides). Note: resistor rules are the opposite of capacitor rules.

EMF and internal resistance: A cell provides EMF $\epsilon$, and the terminal voltage is V = $\epsilon$ - Ir, where r is internal resistance.
As current increases, terminal voltage decreases; [$\epsilon$] = [M $L^{2} \; T^{-3} \; A^{-1}$], SI unit: volt.

Kirchhoff's laws are the foundation of circuit analysis. The junction rule (sum of currents at a junction = 0) follows from charge conservation. The loop rule (sum of potential changes around a closed loop = 0) follows from energy conservation. Sign convention: traversing a resistor in the direction of current gives -IR; traversing a cell from - to + gives +$\epsilon$.

The Wheatstone bridge is balanced when P/Q = R/S, giving zero galvanometer current. At balance, the galvanometer arm can be removed without affecting the circuit. The metre bridge is a practical Wheatstone bridge: R/S = l/(100 - l), where l is the balance length (cm).

The potentiometer uses the principle that potential drop across a uniform wire is proportional to length (V proportional to l for constant current). It measures true EMF because at the null point, no current flows through the test cell.
EMF comparison: $\frac{\epsilon_{1}}{\epsilon_{2}}$ = $\frac{l_{1}}{l_{2}}$.
Internal resistance measurement: r = R($l_{1}$ - $l_{2}$)/$l_{2}$.

The key testable concept is Kirchhoff's laws applied to Wheatstone bridge and potentiometer problems, which together contribute the majority of numerical questions from this chapter in NEET.

Solved Numericals

N₁. A copper wire of length 2 m and cross-sectional area 1 $mm^{2}$ carries a current of 1 A. Given n = 8.5 x $10^{28}$ /$m^{3}$ for copper, find the drift velocity.

Given: l = 2 m, A = 1 $mm^{2}$ = 1 x $10^{-6} \; m^{2}$, I = 1 A, n = 8.5 x $10^{28}$ /$m^{3}$. Using I = neAvd: vd = I/(neA) = 1 A / (8.5 x $10^{28}$ /$m^{3}$ x 1.6 x $10^{-19}$ C x 1 x $10^{-6} \; m^{2}$) vd = 1 / (8.5 x $10^{28}$ x 1.6 x $10^{-19}$ x $10^{-6}$) m/s vd = 1 / (8.5 x 1.6 x $10^{3}$) m/s vd = 1 / (13600) m/s vd = 7.35 x $10^{-5}$ m/s ~ 0.074 mm/s.

This confirms drift velocity is extremely small. At this speed, an electron would take 2 m / (7.35 x $10^{-5}$ m/s) = 2.72 x $10^{4}$ s ~ 7.5 hours to traverse the wire, yet current flows instantly because the electric field signal propagates at nearly c.

N₂. In a Wheatstone bridge, P = 100 ohm, Q = 150 ohm, R = 40 ohm. Find S for balanced condition. If the galvanometer and battery are interchanged, is the bridge still balanced?

Balance condition: P/Q = R/S. $\frac{100}{150}$ = 40/S S = 40 x $\frac{150}{100}$ = 60 ohm.

When galvanometer and battery are interchanged, the balance condition P/Q = R/S remains the same (this is the reciprocity theorem for Wheatstone bridge). Therefore, the bridge is still balanced with S = 60 ohm.

N₃. A potentiometer wire of length 1 m has a resistance of 10 ohm. It is connected in series with a cell of EMF 2 V (internal resistance 1 ohm) and an external resistance $R_{ext}$. A cell of EMF 1.2 V balances at 60 cm. Find the external resistance.

Current through the potentiometer wire: I = $\epsilon$/($R_{wire}$ + r + $R_{ext}$) = 2/(10 + 1 + $R_{ext}$) A. Potential drop across the wire: $V_{wire}$ = I x $R_{wire}$ = 2 x 10/(11 + $R_{ext}$) V. Potential gradient: k = $V_{wire}$/L = 20/(11 + $R_{ext}$) V/m. At balance, the EMF of the test cell equals the potential drop across balance length: $\epsilon_{cell}$ = k x $l_{balance}$ 1.2 = [20/(11 + $R_{ext}$)] x 0.60 1.2 = 12/(11 + $R_{ext}$) 11 + $R_{ext}$ = $\frac{12}{1}$.2 = 10. $R_{ext}$ = 10 - 11 = -1.

This gives a negative value, which is physically impossible. Let us re-examine. The potential drop across the full wire must be greater than or equal to the EMF of the cell being balanced. $V_{wire}$ = 20/(11 + $R_{ext}$).
For $V_{wire}$ >= 1.2 V, we need 11 + $R_{ext}$ <= 16.67, so $R_{ext}$ <= 5.67 ohm. At balance at 60 cm: 1.2 = [20/(11 + $R_{ext}$)] x 0.60 1.2(11 + $R_{ext}$) = 12 13.2 + 1.2 $R_{ext}$ = 12 1.2 $R_{ext}$ = -1.2.

The arithmetic indicates $R_{ext}$ should be reconsidered.
The issue is that with the given parameters, the potential gradient across 60 cm with just the wire resistance of 10 ohm and internal resistance of 1 ohm (no external resistance) gives: V across 60 cm = [2/(10+1)] x 10 x 0.6 = 2 x 10 x 0.$\frac{6}{11}$ = $\frac{12}{11}$ = 1.09 V, which is less than 1.2 V. Hence no balance is possible without reducing total series resistance. Re-reading: the external resistance must be such that we get higher potential gradient.
Actually if $R_{ext}$ = 0, V across 60 cm = ($\frac{2}{11}$) x 10 x ($\frac{60}{100}$) = 1.09 V < 1.2 V. For balance at 60 cm with 1.2 V, we need the driver cell EMF to be higher, or the wire resistance distribution to be different.

Corrected approach: The balance condition requires a higher potential gradient. With the given values, balance would occur if the wire resistance per cm is recalculated. Let us use: potential at 60 cm = (resistance of 60 cm / total circuit resistance) x EMF. Resistance of 60 cm = 10 x 0.6 = 6 ohm. At null: 1.2 = I x 6 = [2/(11 + $R_{ext}$)] x 6. 1.2(11 + $R_{ext}$) = 12 → 11 + $R_{ext}$ = 10.

Since total resistance cannot be less than wire + internal = 11 ohm, this means $R_{ext}$ is not needed and the balance cannot occur at 60 cm with these exact parameters. In the exam, this would indicate $R_{ext}$ = 0 and the balance length is recalculated: l = 1.2 x 11 x 100/(2 x 10) = 66 cm. This is a common trick: check if the parameters are self-consistent before solving.

For the intended solution type: If the driver EMF were 3 V (r = 1 ohm), then 1.2 = [3/(11 + $R_{ext}$)] x 6 gives 11 + $R_{ext}$ = 15, so $R_{ext}$ = 4 ohm. This demonstrates the importance of checking parameter consistency in potentiometer problems.