Current Electricity — Ohm's Law, Kirchhoff's Laws, Circuits

The Flow of Charge

Current electricity is about charges in motion. Unlike electrostatics (charges at rest), here we study what happens when charges flow through conductors — how current depends on voltage, how resistances combine, and how to analyze complex circuits using Kirchhoff’s laws.

This is a high-weightage chapter at every level. CBSE Class 10 dedicates 5-6 marks, Class 12 boards give 7-8 marks, and JEE Main tests 1-2 questions per session.

graph TD
    A[Circuit Problem] --> B{What type?}
    B -->|Simple series/parallel| C[Combine resistances]
    B -->|Complex network| D[Kirchhoff's Laws]
    B -->|Wheatstone bridge| E[Balance condition]
    B -->|Internal resistance| F[EMF = V + Ir]
    C --> G{Series?}
    G -->|Yes| H[R = R₁ + R₂ + ...]
    G -->|Parallel| I[1/R = 1/R₁ + 1/R₂ + ...]
    D --> J[KCL: ΣI = 0 at node]
    D --> K[KVL: ΣV = 0 in loop]
    E --> L["R₁/R₂ = R₃/R₄ → Ig = 0"]

Key Terms & Definitions

Electric Current — Rate of flow of charge: $I = Q/t$ . SI unit: ampere (A).

Potential Difference (Voltage) — Work done per unit charge in moving charge between two points: $V = W/Q$ . Unit: volt (V).

Resistance — Opposition to current flow: $R = V/I$ . Unit: ohm ( $\Omega$ ).

EMF — Electromotive force. The potential difference across a cell when no current flows. Not actually a force — it’s energy per unit charge provided by the cell.

Internal Resistance — Resistance of the cell itself. Terminal voltage $V = \varepsilon - Ir$ where $\varepsilon$ is EMF and $r$ is internal resistance.

Essential Formulas

V = IR

R = \rho \frac{L}{A}

where $\rho$ is resistivity (material property), $L$ is length, $A$ is cross-sectional area.

Series: $R_{eq} = R_1 + R_2 + \ldots$ (same current, voltages add)

Parallel: $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \ldots$ (same voltage, currents add)

For two resistors in parallel: $R_{eq} = \frac{R_1 R_2}{R_1 + R_2}$

Junction Rule (KCL): Sum of currents entering a junction = sum of currents leaving.

Loop Rule (KVL): Sum of potential differences around any closed loop = 0.

P = VI = I^2R = \frac{V^2}{R}

H = I^2Rt \quad \text{(Joule's law of heating)}

Balanced when: $\frac{R_1}{R_2} = \frac{R_3}{R_4}$

No current flows through the galvanometer when balanced. Used for precise resistance measurement.

Solved Examples — Easy to Hard

Example 1 (Easy — CBSE Class 10)

Three resistors of 2 $\Omega$ , 3 $\Omega$ , and 6 $\Omega$ are in parallel. Find equivalent resistance.

\frac{1}{R} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3+2+1}{6} = 1 \implies R = \mathbf{1\ \Omega}

Example 2 (Medium — CBSE Class 12)

A cell of EMF 1.5 V and internal resistance 0.5 $\Omega$ is connected to a 2.5 $\Omega$ resistor. Find current and terminal voltage.

$I = \frac{\varepsilon}{R + r} = \frac{1.5}{2.5 + 0.5} = \frac{1.5}{3} = \mathbf{0.5 \text{ A}}$

$V = \varepsilon - Ir = 1.5 - 0.5(0.5) = \mathbf{1.25 \text{ V}}$

Example 3 (Hard — JEE Main)

In a Wheatstone bridge, $R_1 = 100\ \Omega$ , $R_2 = 200\ \Omega$ , $R_3 = 150\ \Omega$ . For balance, find $R_4$ .

Balance: $R_1/R_2 = R_3/R_4$

$R_4 = R_3 \times R_2/R_1 = 150 \times 200/100 = \mathbf{300\ \Omega}$

Exam-Specific Tips

CBSE Class 10: Series-parallel combination questions and Joule heating are the staples. Know the formulas cold and practise the “equivalent resistance” problems.

CBSE Class 12: Kirchhoff’s laws derivation and application, Wheatstone bridge, and meter bridge problems carry high marks. Potentiometer is also important — comparison of EMFs and finding internal resistance.

JEE Main: Complex circuit analysis using Kirchhoff’s laws, symmetry in resistor networks, and temperature dependence of resistance are frequently tested.

Common Mistakes to Avoid

Mistake 1 — Adding resistances in parallel like series. In parallel, you add reciprocals, not resistances directly. The equivalent is always less than the smallest individual resistance.

Mistake 2 — Ignoring internal resistance. When a problem mentions a “cell with EMF $\varepsilon$ and internal resistance $r$ ”, the terminal voltage is $\varepsilon - Ir$ , not just $\varepsilon$ .

Mistake 3 — Wrong sign convention in KVL. Going from $+$ to $-$ through a cell is a voltage drop. Going against current through a resistor is a voltage rise. Stick to one convention.

Mistake 4 — Confusing EMF with terminal voltage. EMF is the “open circuit” voltage. When current flows, terminal voltage is less than EMF due to internal resistance.

Mistake 5 — Forgetting power dissipation in internal resistance. Total power from cell = $\varepsilon I$ . Power to external circuit = $I^2 R$ . Power lost inside = $I^2 r$ .

Practice Questions

Q1. Find equivalent resistance of 4 $\Omega$ and 6 $\Omega$ in parallel.

$R = (4 \times 6)/(4+6) = 24/10 = 2.4\ \Omega$ .

Q2. A wire of resistance 10 $\Omega$ is stretched to double its length. Find new resistance.

Length doubles, area halves (volume constant). $R' = \rho(2L)/(A/2) = 4\rho L/A = 40\ \Omega$ .

Q3. Current of 2 A flows through a 5 $\Omega$ resistor for 10 min. Find heat produced.

$H = I^2Rt = 4 \times 5 \times 600 = 12000$ J $= 12$ kJ.

Q4. A potentiometer balances a 1.5 V cell at 75 cm. Where will a 1.2 V cell balance?

$l_2/l_1 = E_2/E_1$ . $l_2 = 75 \times 1.2/1.5 = 60$ cm.

Q5. In a meter bridge, null point is at 40 cm. If known resistance is 12 $\Omega$ , find unknown.

$R/S = l/(100-l)$ . $R/12 = 40/60$ . $R = 8\ \Omega$ .

Q6. Maximum power transfer: for what external resistance $R$ is power maximum from a cell of EMF $\varepsilon$ and internal resistance $r$ ?

$P = I^2 R = \varepsilon^2 R/(R+r)^2$ . Differentiate and set to zero: $R = r$ . Maximum power when external resistance equals internal resistance.

Q7. 12 resistors of 1 $\Omega$ each form the edges of a cube. Find resistance between opposite corners.

By symmetry, the effective resistance between diagonally opposite corners of a resistor cube is $5/6\ \Omega$ .

Q8. A 100 W bulb and a 60 W bulb are connected in series to 220 V. Which glows brighter?

Rated power tells us resistance: $R = V^2/P$ . $R_{100} = 484\ \Omega$ , $R_{60} = 806.7\ \Omega$ . In series, same current flows. Power $= I^2R$ , so the higher-resistance (60 W rated) bulb dissipates more power and glows brighter.

FAQs

What is the difference between EMF and potential difference?

EMF is the total energy per unit charge supplied by the source (measured in open circuit). Potential difference is the energy per unit charge between any two points when current flows. Due to internal resistance, terminal PD is less than EMF.

Why do resistances add in series but reciprocals add in parallel?

In series, same current passes through each — voltages add. $V = V_1 + V_2 = IR_1 + IR_2 = I(R_1+R_2)$ . In parallel, same voltage across each — currents add. $I = V/R_1 + V/R_2$ , giving $1/R = 1/R_1 + 1/R_2$ .

What is drift velocity?

The average velocity of free electrons in a conductor due to an applied field. It’s very slow (~mm/s) even though current “travels” at near light speed (because the field propagates fast).

How does temperature affect resistance?

For metals, resistance increases with temperature: $R = R_0(1 + \alpha T)$ . For semiconductors, resistance decreases with temperature.

What is a superconductor?

A material with zero resistance below a critical temperature. Current flows without any energy loss. Used in MRI machines and particle accelerators.

Advanced Concepts

Meter bridge — practical Wheatstone bridge

A meter bridge is a 1 m wire of uniform cross-section. An unknown resistance $R$ and a known resistance $S$ are connected as two arms. A galvanometer finds the null point at length $l$ :

\frac{R}{S} = \frac{l}{100 - l}

CBSE board exam: “Describe the working of a meter bridge.” This is a 5-mark question. Draw the diagram, derive the balance condition, and mention that the wire must be uniform and the null point should be near the centre for best accuracy.

Potentiometer — the precision instrument

A potentiometer measures EMF without drawing current (unlike a voltmeter). A long wire of uniform cross-section carries a steady current. The potential drop per unit length is constant.

Comparing two EMFs: $\frac{\varepsilon_1}{\varepsilon_2} = \frac{l_1}{l_2}$

Finding internal resistance: Connect cell with external resistance $R$ : $r = R\left(\frac{l_1 - l_2}{l_2}\right)$

where $l_1$ is the balance length without $R$ and $l_2$ is with $R$ .

Drift velocity and current density

I = nAv_d e

where $n$ = free electron density, $A$ = cross-section area, $v_d$ = drift velocity, $e$ = electron charge.

Current density: $J = I/A = nv_d e = \sigma E$

where $\sigma = 1/\rho$ is the conductivity.

Drift velocity is very small ( $\sim 10^{-4}$ m/s in copper) — but the electric field propagates at near light speed, which is why a bulb lights up instantly when you flip the switch.

Symmetry in resistor networks

For complex networks, look for symmetry:

Wheatstone balance: If the bridge is balanced, the galvanometer arm carries zero current — remove it.
Equipotential points: If two points are at the same potential by symmetry, they can be shorted together without changing any current.
Cube of resistors: By symmetry, the 12-resistor cube has effective resistance $5R/6$ between opposite corners.

Additional Solved Example

A copper wire has resistance 10 $\Omega$ at 20°C. Find its resistance at 120°C. (Temperature coefficient $\alpha = 3.9 \times 10^{-3}$ /°C)

$R = R_0(1 + \alpha\Delta T) = 10(1 + 3.9 \times 10^{-3} \times 100) = 10(1.39) = 13.9\,\Omega$

Additional Practice Questions

Q9. A potentiometer wire is 10 m long. A cell of EMF 2 V balances at 8 m. Find the potential gradient.

Potential gradient = EMF / balance length is NOT correct here. The potential gradient along the wire = $V_{driving}/\text{total length}$ . Since 2 V balances at 8 m, the potential gradient = $2/8 = 0.25$ V/m.

Q10. The drift velocity of electrons in a copper wire carrying 1 A of current is $v$ . If the current is increased to 2 A, what is the new drift velocity?

$I = nAv_d e$ . Since $n$ , $A$ , and $e$ are constant, $v_d \propto I$ . New drift velocity = $2v$ .