ICSE Class 10 Physics: Ohm's Law & Simple Circuit Problems

Ever wondered why a tiny battery can light up a whole bulb? The secret lies in a simple rule that every electrician and physics student swears by.

💡 In Simple Words: Ohm's law says that the electric push (voltage) across a component makes a flow of tiny particles (current) move through it, and the amount of flow depends on how hard the component resists (resistance). Double the push, double the flow – as long as the resistance stays the same.

What is Ohm's Law?

Ohm's law is a relationship between three things: voltage (V), current (I), and resistance (R). Think of voltage like water pressure in a hose, current like the amount of water flowing, and resistance like a narrow section of the hose that slows the flow down.

The formula is:

V = I × R

That means if you know any two of the three, you can find the third. It’s the backbone of almost every problem you’ll see in the ICSE electricity chapter.

Key Terms Explained

• Voltage (V): The electric push that makes charges move, measured in volts. Imagine it as the height difference that makes water flow downhill.
• Current (I): The rate at which electric charge flows, measured in amperes (amps). It’s like the amount of water that passes a point each second.
• Resistance (R): Anything that opposes the flow of charge, measured in ohms (Ω). A thin pipe offers more resistance than a wide one.

Solving Simple Circuit Problems

Most exam questions give you two of the three values and ask you to find the third. The trick is to stay organized and avoid unit mix‑ups.

Step‑by‑Step Method (with a handy flowchart)

Follow these four steps every time you see a circuit problem:

Worked Example 1: Finding Current

Problem: A 9 V battery is connected to a resistor of 3 Ω. What is the current flowing?

Solution:

1. Write the formula: V = I × R.
2. Plug in the known numbers: 9 = I × 3.
3. Solve for I: I = 9 ÷ 3 = 3 A.

So, the current is 3 amperes. Easy, right?

Worked Example 2: Finding Resistance

Problem: A lamp draws 0.5 A when connected to a 12 V supply. What is its resistance?

Solution:

1. Start with V = I × R.
2. Rearrange to R = V ÷ I.
3. Insert the values: R = 12 ÷ 0.5 = 24 Ω.

The lamp’s resistance is 24 ohms.

Worked Example 3: Series Circuit

In a series circuit, resistors are linked end‑to‑end, so the same current flows through each. The total resistance is just the sum of individual resistances.

Problem: Two resistors, 5 Ω and 10 Ω, are connected in series to a 15 V battery. Find the current.

Solution:

1. Total resistance Rtotal = 5 + 10 = 15 Ω.
2. Use V = I × Rtotal: 15 = I × 15.
3. I = 15 ÷ 15 = 1 A.

The circuit carries 1 ampere of current.

Worked Example 4: Parallel Circuit

In a parallel circuit, each branch gets the full voltage, but the total current is the sum of branch currents. The reciprocal of total resistance equals the sum of reciprocals of each branch resistance.

Problem: Two resistors, 6 Ω and 3 Ω, are connected in parallel across a 12 V source. Find the total current supplied by the battery.

Solution:

1. Calculate the equivalent resistance: 1/Req = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 → Req = 2 Ω.
2. Now use V = I × Req: 12 = I × 2.
3. I = 12 ÷ 2 = 6 A.

The battery provides 6 amperes in total.

Quick Reference Table

Common Mistakes to Avoid

• Mixing up units – always keep volts, amperes, and ohms separate.
• Forgetting that in series circuits the same current flows through each resistor.
• Ignoring that in parallel circuits each branch sees the full battery voltage.
• Not checking if the answer makes sense (e.g., a huge current from a tiny battery is a red flag).

📝 Likely Exam Questions

1. Question: A 6 Ω resistor is connected to a 9 V cell. Calculate the current.
2. Answer: I = V ÷ R = 9 ÷ 6 = 1.5 A.
3. Question: Two resistors of 4 Ω and 6 Ω are connected in series across a 20 V supply. Find the voltage across each resistor.
4. Answer: Total R = 10 Ω, total I = 20 ÷ 10 = 2 A. Voltage across 4 Ω = I × R = 2 × 4 = 8 V; across 6 Ω = 2 × 6 = 12 V.
5. Question: A parallel circuit has two branches: one 12 Ω, the other 8 Ω, connected to a 24 V battery. Determine the total current supplied.
6. Answer: 1/Req = 1/12 + 1/8 = 1/12 + 3/24 = 5/24 → Req = 24/5 ≈ 4.8 Ω. Itotal = V ÷ Req = 24 ÷ 4.8 = 5 A.
7. Question: Explain why the current is the same through each resistor in a series circuit but the voltage drops differ.
8. Answer: In series, the same charge passes through each component, so the flow (current) is identical. Voltage divides according to each resistor’s value because V = I × R, and different R give different V.