The main objective of this question is to find the** power dissipated** in **each bulb** that is **connected** in **series**.

This question uses the concept of **power in series**. In a **series circuit**, the total** power** is the **same** as the **total** amount of** power lost** by **each resistor**. **Mathematically**, it is **represented** as:

\[ \space P_T \space = \space P_1 \space + \space P_2 \space + \space P_3 \]

**Where** $P_T $ is the total power.

## Expert Answer

**Given** that:

\[ \space R_1 \space = \space 400 \space ohm \]

\[ \space R_1 \space = \space 800 \space ohm \]

**Voltage** is:

\[ \space V \space = \space 1 2 0 \space V \]

We **know** that:

\[ \space P \space = \space \frac{V^2}{R} \]

So, for the **first bulb**, we have:

\[ \space P_1 \space = \space \frac{V^2}{R_1} \]

By **putting** in the values, we get:

\[ \space P_1 \space = \space \frac{1 2 0^2}{4 0 0} \]

\[ \space P_1 \space = \space \frac{1 4 4 0 0}{4 0 0} \]

\[ \space P_1 \space = \space 3 6 \space W \]

Now for the **second bulb**, we have:

\[ \space P_2 \space = \space \frac{V^2}{R_2} \]

By **putting** in the **values**, we get:

\[ \space P_1 \space = \space \frac{1 2 0^2}{8 0 0} \]

\[ \space P_1 \space = \space \frac{1 4 4 0 0}{8 0 0} \]

\[ \space P_1 \space = \space 1 8 \space W \]

## Numerical Answer

The **power dissipated** in the **first bulb** is:

\[ \space P_1 \space = \space 3 6 \space W \]

**And** for the **second bulb**, the **power dissipated** is:

\[ \space P_1 \space = \space 1 8 \space W \]

## Example

In the **above question**, if the r**esistance** across **one bulb** is $ 600 $ **ohm** and 1200 **ohm** across **another bulb**. Find the **power dissipated** along these **two bulbs** which are **connected** in **series**.

**Given** that:

\[ \space R_1 \space = \space 6 0 0 \space ohm \]

\[ \space R_1 \space = \space 1 2 0 0 \space ohm \]

**Voltage** is:

\[ \space V \space = \space 1 2 0 \space V \]

We **know** that:

\[ \space P \space = \space \frac{V^2}{R} \]

So, for the **first bulb**, we have:

\[ \space P_1 \space = \space \frac{V^2}{R_1} \]

By **putting** in the values, we get:

\[ \space P_1 \space = \space \frac{1 2 0^2}{6 0 0} \]

\[ \space P_1 \space = \space \frac{1 4 4 0 0}{6 0 0} \]

\[ \space P_1 \space = \space 24 \space W \]

Now for the **second bulb**, we have:

\[ \space P_2 \space = \space \frac{V^2}{R_2} \]

By **putting** in the **values**, we get:

\[ \space P_1 \space = \space \frac{1 2 0^2}{1 2 0 0} \]

\[ \space P_1 \space = \space \frac{1 4 4 0 0}{1 2 0 0} \]

\[ \space P_1 \space = \space 1 2 \space W \]

Thus, the **power dissipated** in the **first bulb** is:

\[ \space P_1 \space = \space 2 4 \space W \]

**And** for the **second bulb**, the **power dissipated** is:

\[ \space P_1 \space = \space 1 2 \space W \]