# Ohm’s Law and in-series components

ohms-lawseries

I am doing some practice problems based on what I learned today in lecture. However, I am stuck on this problem:

`A 10Ω resistor is in series with a bulb and a 12V source.`

```a. If 8V is across the bulb, what is across the resistor? b. What is the current in the circuit? c. What is the resistance of the bulb? ```

Where would I begin solving what voltage is across the resistor? I do not the current, nor do I have the means to find it (that I can see), thus I can not find what voltage is dropped. I am not sure if I can assume the bulb has no resistance, in which case I can solve the problem easily.

#### Best Answer

You need to use KVL around the loop:

simulate this circuit – Schematic created using CircuitLab

KVL says that the sum of the voltages around the loop equals 0. That means that if there is a 12V increase traveling from the negative to the positive terminal of the voltage source, then the voltage must fall by 12V traveling through the bulb and resistor.

The current in the loop can be found using Ohm's Law and from the voltage \\$V\\$ across the \\$10\Omega\\$ resistor you found in part (a).

The resistance of the bulb can then be found using Ohm's Law again: you are given the fact that 8V is across the bulb and from part (b) you know the current through it. Solve for the resistance \\$R\\$.