Electricity — electric current, circuit symbols, Ohm's law introduction

Question

A 12 V battery is connected to a resistor of 4 ohms. Find the current flowing through the circuit. What happens to the current if the resistance is doubled?

(CBSE Class 7-10 — Electricity)

Circuit Analysis Basics

flowchart TD
    A["Electric Circuit Problem"] --> B{What is given?}
    B -->|V and R| C["Find I: Ohm's Law"]
    B -->|V and I| D["Find R: R = V/I"]
    B -->|I and R| E["Find V: V = IR"]
    C --> F["I = V/R"]
    A --> G["Key Relations"]
    G --> G1["Series: R_total = R1 + R2"]
    G --> G2["Parallel: 1/R_total = 1/R1 + 1/R2"]
    G --> G3["Power: P = VI = I²R = V²/R"]

Solution — Step by Step

Ohm’s law: The current through a conductor is directly proportional to the voltage and inversely proportional to the resistance.

V = IR \implies I = \frac{V}{R}

Given: $V = 12$ V, $R = 4\;\Omega$

I = \frac{12}{4} = \mathbf{3 \text{ A}}

If $R$ is doubled to $8\;\Omega$ (with the same battery):

I_{\text{new}} = \frac{12}{8} = 1.5 \text{ A}

The current halves when resistance doubles. This is the inverse proportionality: $I \propto 1/R$ (at constant voltage).

Electric current is the flow of charge. $I = Q/t$ where $Q$ is charge in coulombs and $t$ is time in seconds. 1 ampere = 1 coulomb per second.

Potential difference (voltage) is the “push” that drives current. A battery creates a potential difference between its terminals.

Resistance opposes current flow. It depends on: material, length (more length = more resistance), cross-section area (thinner wire = more resistance), and temperature.

Why This Works

Ohm’s law is the fundamental relationship in circuit analysis. Think of water flowing through a pipe: voltage is the water pressure, current is the flow rate, and resistance is the pipe’s narrowness. Higher pressure (voltage) pushes more water (current). A narrower pipe (higher resistance) allows less flow.

This hydraulic analogy helps visualise why $I = V/R$ : more push and less obstruction means more flow.

Alternative Method — Using Power

We can also find the power dissipated:

P = VI = 12 \times 3 = 36 \text{ W}

Or equivalently: $P = I^2R = 9 \times 4 = 36$ W, or $P = V^2/R = 144/4 = 36$ W.

All three forms give the same answer — use whichever has the given quantities.

For CBSE Class 10, you need to solve series and parallel circuit problems. Key facts: in series, current is the same through all resistors and voltages add up. In parallel, voltage is the same across all resistors and currents add up. Draw the circuit first, label all known values, then apply Ohm’s law to each part.

Common Mistake

The most common error: confusing series and parallel formulas. In series, resistances add directly ( $R_{\text{total}} = R_1 + R_2$ ). In parallel, reciprocals add ( $1/R_{\text{total}} = 1/R_1 + 1/R_2$ ). Students often swap these. A quick check: in parallel, the total resistance is always LESS than the smallest individual resistance. If your answer violates this, you have used the wrong formula.