Why Ohm’s Law matters in everyday life
Ever wondered why a flashlight dims when you swap batteries? That tiny change is all about Ohm’s Law, the rule that ties voltage, current, and resistance together.
💡 In Simple Words: Ohm’s Law says the electric push (voltage) equals how much flow (current) you get multiplied by how hard the path resists (resistance). Change one, and the others adjust.
What is Ohm’s Law?
Voltage (V) is the electric pressure that pushes charges through a wire, much like water pressure pushes water through a pipe. Current (I) is the rate at which charge flows, similar to the amount of water flowing per second. Resistance (R) is anything that slows that flow down, like a narrow section of pipe. The law itself is written as V = I × R. If you know any two of these, you can find the third.
How to use Ohm’s Law in simple circuits
Most exam questions give you a tiny circuit with a battery and a few resistors. Follow these steps:
- Identify all the resistors and how they are connected (series or parallel).
- Combine resistors to get an equivalent resistance.
- Apply V = I R using the battery’s voltage to find the total current.
- Use the current to find voltage drops across individual resistors (V = I R again, but with the resistor’s own value).
Series vs. Parallel: quick comparison
| Feature | Series connection | Parallel connection |
|---|---|---|
| Current through each resistor | Same current flows through all | Current splits, each branch gets its own share |
| Voltage across each resistor | Divides proportionally to resistance | Same voltage across all branches |
| Total resistance | R_total = R1 + R2 + … | 1/R_total = 1/R1 + 1/R2 + … |
| Effect of adding another resistor | Increases total resistance | Decreases total resistance |
Worked example 1: Find the current
A 12 V battery is connected to two resistors, 3 Ω and 6 Ω, in series. What is the current flowing through the circuit?
- Because the resistors are in series, add them: R_total = 3 Ω + 6 Ω = 9 Ω.
- Apply Ohm’s Law: I = V / R = 12 V / 9 Ω = 1.33 A (approximately).
The same 1.33 A flows through both resistors.
Worked example 2: Voltage drop across each resistor
Using the same circuit, find the voltage across the 3 Ω resistor.
- V_drop = I × R = 1.33 A × 3 Ω ≈ 4 V.
- The remaining 8 V appears across the 6 Ω resistor (12 V total – 4 V).
Worked example 3: A mixed circuit
Three resistors are connected as follows: R1 = 4 Ω and R2 = 6 Ω are in parallel, and this combination is in series with R3 = 5 Ω. The battery provides 24 V. Find the total current.
- First find the parallel equivalent: 1/R_parallel = 1/4 + 1/6 = (3+2)/12 = 5/12 → R_parallel = 12/5 = 2.4 Ω.
- Add the series resistor: R_total = 2.4 Ω + 5 Ω = 7.4 Ω.
- Current from the battery: I_total = V / R_total = 24 V / 7.4 Ω ≈ 3.24 A.
Now you can also find the current through each branch of the parallel part using I = V / R for that part (the voltage across the parallel network is the same as across R3, i.e., 24 V – I_total×5 Ω ≈ 24 V – 3.24×5 ≈ 8.8 V).
Quick tip sheet
- Always write down what you know: V, I, or R.
- Remember the series‑parallel rules – they’re the backbone of every circuit problem.
- Check units: volts (V), amperes (A), ohms (Ω). Mis‑matching units is a common slip.
- When in doubt, draw the circuit again, label each resistor, and mark the direction of current.
📝 Likely Exam Questions
- Question: A 9 V battery is connected to three resistors of 2 Ω, 3 Ω, and 6 Ω all in series. Find the current and the voltage across the 3 Ω resistor.
Answer: R_total = 11 Ω, I = 9 V / 11 Ω = 0.82 A, V_3Ω = 0.82 A × 3 Ω = 2.46 V. - Question: Two resistors, 5 Ω and 10 Ω, are connected in parallel across a 15 V source. Determine the total resistance and the total current supplied by the source.
Answer: 1/R_total = 1/5 + 1/10 = 0.3 → R_total = 3.33 Ω, I_total = 15 V / 3.33 Ω ≈ 4.5 A. - Question: In a circuit, a 12 V battery feeds a series combination of a 4 Ω resistor and a parallel network of 8 Ω and 12 Ω. Find the current through the 4 Ω resistor.
Answer: Parallel: 1/R_p = 1/8 + 1/12 = 5/24 → R_p = 4.8 Ω. Total R = 4 Ω + 4.8 Ω = 8.8 Ω. I_total = 12 V / 8.8 Ω ≈ 1.36 A. This is the current through the 4 Ω resistor. - Question: Explain why the current is the same through each resistor in a series circuit but the voltage can be different.
Answer: In series, there is only one path for charge, so the same amount of charge per second (current) must flow through every component. Voltage, however, is the energy per charge; each resistor uses a portion of the total voltage proportional to its resistance (V = I R). - Question: A student measures 2 A flowing through a resistor of unknown value when a 10 V battery is connected. What is the resistance?
Answer: R = V / I = 10 V / 2 A = 5 Ω.