Current Electricity | CBSE - Wyatt's Notes

sources:

• text: Standard textbook reference

Current electricity deals with the flow of electric charge through conductors. It covers Ohm’s law, Kirchhoff’s laws, series and parallel circuits, and electrical measurements.

• Ohm’s law: $V = IR$ (for ohmic conductors at constant temperature)
• Resistance: $R = \rho\frac{L}{A}$ where $\rho$ is resistivity, $L$ is length, $A$ is cross-sectional area
• Series circuits: $R_{total} = R_1 + R_2 + ...$, current is same through all components
• Parallel circuits: $\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + ...$, voltage is same across all components
• Kirchhoff’s junction rule: $\sum I_{in} = \sum I_{out}$ (conservation of charge)
• Kirchhoff’s loop rule: $\sum V = 0$ around any closed loop (conservation of energy)
• Power: $P = IV = I^2R = \frac{V^2}{R}$

Problem: Three resistors of 2 $\Omega$, 3 $\Omega$, and 5 $\Omega$ are connected in series to a 10 V battery. Find the current through each resistor and the voltage across each.

Solution:

Total resistance: $R_{total} = 2 + 3 + 5 = 10 \, \Omega$

Current (same through all): $I = \frac{V}{R_{total}} = \frac{10}{10} = 1 \, \text{A}$

Voltages: $V_1 = IR_1 = 1 \times 2 = 2 \, \text{V}$ $V_2 = IR_2 = 1 \times 3 = 3 \, \text{V}$ $V_3 = IR_3 = 1 \times 5 = 5 \, \text{V}$

Check: $V_1 + V_2 + V_3 = 2 + 3 + 5 = 10 \, \text{V}$ (matches the battery voltage).

Common mistake: Assuming the voltage is the same across all resistors in series. In series, current is constant, not voltage.

Problem: Two resistors of 6 $\Omega$ and 3 $\Omega$ are connected in parallel to a 12 V battery. Find the current through each resistor and the total current.

Solution:

Voltage across each resistor is 12 V (parallel circuit).

Current through 6 $\Omega$: $I_1 = \frac{V}{R_1} = \frac{12}{6} = 2 \, \text{A}$

Current through 3 $\Omega$: $I_2 = \frac{V}{R_2} = \frac{12}{3} = 4 \, \text{A}$

Total current: $I_{total} = I_1 + I_2 = 2 + 4 = 6 \, \text{A}$

Equivalent resistance: $R_{eq} = \frac{V}{I_{total}} = \frac{12}{6} = 2 \, \Omega$

Common mistake: Forgetting that current splits in parallel circuits. The total current is the sum of branch currents.

Problem: A Wheatstone bridge has resistors $R_1 = 100 \, \Omega$, $R_2 = 200 \, \Omega$, $R_3 = 150 \, \Omega$ in three arms. Find the value of $R_4$ for the bridge to be balanced.

Solution:

For a balanced Wheatstone bridge: $\frac{R_1}{R_2} = \frac{R_3}{R_4}$

$R_4 = \frac{R_2 \times R_3}{R_1} = \frac{200 \times 150}{100} = 300 \, \Omega$

Common mistake: Getting the ratio order wrong. The resistors must be in opposite arms of the bridge.

1. A 12 V battery is connected to two resistors (4 $\Omega$ and 6 $\Omega$) in parallel. Find the total current and power dissipated in each resistor.
2. Find the equivalent resistance of three resistors (2 $\Omega$, 3 $\Omega$, 6 $\Omega$) connected in parallel.
3. A potentiometer wire of length 1 m has resistance 10 $\Omega$. Find the balancing length when a cell of EMF 1.5 V is balanced against a standard cell of 2 V.

Electric current is charge in motion — like water flowing through pipes: Think of voltage as water pressure, current as flow rate, and resistance as pipe narrowness. Ohm’s law (V = IR) is the electrical equivalent of “more pressure pushes more water through a narrow pipe.” Series circuits are like pipes connected end-to-end — the same water flows through each section. Parallel circuits are like branching pipes — the flow splits among branches, each getting the same pressure but different amounts of water depending on their resistance.

Why it matters: Current electricity is literally the lifeblood of modern civilization — it powers everything from lighting to computing to communication. Understanding circuits means understanding how to design, build, and troubleshoot the electrical systems that run our world.

The key insight: Kirchhoff’s laws are just conservation laws in disguise — the junction rule conserves charge (what goes in must come out), and the loop rule conserves energy (what you gain going around must equal what you lose).

• Draw the circuit diagram before solving
• Identify series and parallel combinations first
• Use Kirchhoff’s laws for complex circuits
• Always check units (ohms, volts, amperes)
• Power calculations often appear in multi-step problems

• Ohm’s law: $V = IR$
• Resistance: $R = \rho\frac{L}{A}$
• Series: $R_{eq} = R_1 + R_2 + ...$, $V_{total} = V_1 + V_2 + ...$, $I$ is constant
• Parallel: $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + ...$, $V$ is constant, $I_{total} = I_1 + I_2 + ...$
• Power: $P = IV = I^2R = \frac{V^2}{R}$
• Internal resistance: $V_{terminal} = E - Ir$
• Temperature dependence: $R = R_0[1 + \alpha(T - T_0)]$

Problem: Find the equivalent resistance of the circuit shown below: $R_1 = 6\,\Omega$ and $R_2 = 3\,\Omega$ are in parallel, and this combination is in series with $R_3 = 4\,\Omega$.

Solution:

Step 1: Parallel combination of $R_1$ and $R_2$: $R_{12} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{6 \times 3}{6 + 3} = \frac{18}{9} = 2\,\Omega$

Step 2: Series combination with $R_3$: $R_{eq} = R_{12} + R_3 = 2 + 4 = 6\,\Omega$

Common mistake: Adding the parallel resistors directly without using the parallel formula. Parallel resistors always give a smaller equivalent resistance.

Problem: In the circuit below, find the current through each resistor. Battery EMF = 12 V, $R_1 = 4\,\Omega$, $R_2 = 6\,\Omega$, $R_3 = 3\,\Omega$. $R_1$ and $R_2$ are in parallel, connected to the battery through $R_3$.

Solution:

Let $I$ be the total current through $R_3$, and $I_1$, $I_2$ be currents through $R_1$ and $R_2$ respectively.

By junction rule: $I = I_1 + I_2$

Voltage across parallel combination: $V_{12} = I_1 R_1 = I_2 R_2$ $4I_1 = 6I_2 \implies I_1 = 1.5 I_2$

Loop rule for outer loop: $E - IR_3 - I_1 R_1 = 0$ $12 - (I_1 + I_2) \times 3 - 4I_1 = 0$

Substitute $I_1 = 1.5I_2$: $12 - (1.5I_2 + I_2) \times 3 - 4(1.5I_2) = 0$ $12 - 7.5I_2 - 6I_2 = 0$ $12 = 13.5I_2$ $I_2 = \frac{12}{13.5} = \frac{8}{9} \approx 0.889\,\text{A}$

$I_1 = 1.5 \times \frac{8}{9} = \frac{4}{3} \approx 1.333\,\text{A}$

$I = I_1 + I_2 = \frac{4}{3} + \frac{8}{9} = \frac{12 + 8}{9} = \frac{20}{9} \approx 2.222\,\text{A}$

Common mistake: Forgetting to include the internal resistance of the battery or the series resistor when applying Kirchhoff’s loop rule.

Problem: A copper wire has resistance $10\,\Omega$ at $20°\text{C}$. What is its resistance at $80°\text{C}$? (Temperature coefficient of copper: $\alpha = 0.00404\,°\text{C}^{-1}$)

Solution:

Using the temperature dependence formula: $R = R_0[1 + \alpha(T - T_0)]$

$R = 10[1 + 0.00404 \times (80 - 20)]$

$R = 10[1 + 0.00404 \times 60]$

$R = 10[1 + 0.2424] = 10 \times 1.2424 = 12.424\,\Omega$

Common mistake: Using the wrong temperature difference. Always use $\Delta T = T_{final} - T_{initial}$, not just $T_{final}$.

1. For complex circuits, simplify step by step: identify parallel combinations first, then add series components
2. When using Kirchhoff’s rules, assign current directions consistently; a negative result means the actual direction is opposite
3. The terminal voltage of a battery is less than its EMF when current flows: $V = E - Ir$
4. Power dissipated in a resistor is always positive: $P = I^2R > 0$
5. For maximum power transfer, the load resistance should equal the internal resistance of the source

In a series circuit, the current through each component is the same, but the voltage divides across components proportionally to their resistance ($V = IR$). Students often assume the battery voltage appears across every resistor, which is only true in parallel circuits. Always apply $V_{total} = V_1 + V_2 + \ldots$ for series circuits and use Ohm’s law to find the voltage across each resistor.

When combining resistors in parallel, the equivalent resistance is always less than the smallest individual resistance. Students sometimes add parallel resistors directly ($R_1 + R_2$) instead of using $1/R_{eq} = 1/R_1 + 1/R_2$. A quick sanity check: if two equal resistors $R$ are in parallel, the equivalent is $R/2$, not $2R$.

In a balanced Wheatstone bridge, the ratio is $\frac{R_1}{R_2} = \frac{R_3}{R_4}$ where $R_1$ and $R_2$ are in one branch and $R_3$ and $R_4$ are in the other. Students frequently write the ratio as $\frac{R_1}{R_3} = \frac{R_2}{R_4}$ or mix up which resistors are paired. The correct pairing is determined by the bridge geometry: opposite arms form the ratio.

• Electrostatics — Current electricity involves the flow of charge, building on the electrostatic concepts of charge and electric fields.
• Magnetic Effects of Current — Electric currents produce magnetic fields, connecting current electricity to electromagnetism.
• Electromagnetic Induction — Changing magnetic fields induce currents, linking magnetism back to current electricity through Faraday’s law.