In a circuit of any connection (series or parallel), the power dissipated in the whole circuit is equal to the sum of the power dissipated in each of the individual resistor.

**Example 1**:

2 identical bulb of resistance 3Ω is connected to an e.m.f. of 12V. Find the power dissipated in the circuit if

a. the bulb is connected in series

b. if the bulb is connected in parallel

**Answer**:

a.

Current pass through the 2 resistors,

$$I=\frac{V}{R}=\frac{(12)}{6}=2A$$

Power of each of the resistor,

$$\begin{array}{l}P={I}^{2}R\\ P={\left(2\right)}^{2}(3)=12W\end{array}$$

Sum of the power,

$$\begin{array}{l}P={P}_{1}+{P}_{2}\\ P=(12W)+(12W)=24W\end{array}$$

b.

Potential difference across the 2 resistor = 12V

Power of each of the resistor, $$\begin{array}{l}P=\frac{{V}^{2}}{R}\\ P=\frac{{\left(12\right)}^{2}}{(3)}=48W\end{array}$$ Sum of the power, $$\begin{array}{l}P={P}_{1}+{P}_{2}\\ P=(48W)+(48W)=96W\end{array}$$

**Example 2**:

A 800W heater is used to heat 250 cm³ of water from 30 to 100°C. What is the minimum time in which this can be done? [Density of water = 1000kg/m³; Specific Heat Capacity of water = 4200J°C

^{-1}kg

^{-1}]

**Answer**:

Energy supply by the heater, E = Pt

Heat energy absorbed by the water, E = mcθ

Let's assume that all the energy supplied by the heater is converted to heat energy and absorbed by the water, hence

$$\begin{array}{l}Pt=mc\theta \\ (800)t=(0.25)(4200)(70)\\ t=\frac{(0.25)(4200)(70)}{(800)}=92s\end{array}$$