Electric Power and Brightness in Series vs Parallel Circuits – HSC Physics

This topic is part of the HSC Physics course under the section Electric Circuits.

HSC Physics Syllabus

investigate quantitatively and analyse the rate of conversion of electrical energy in components of electric circuits, including the production of heat and light, by applying `P = VI` and `E = Pt` and variations that involve Ohm’s Law (ACSPH042)

investigate quantitatively the application of the law of conservation of energy to the heating effects of electric currents, including the application of `P = VI` and variations of this involving Ohm’s Law (ACSPH043)

Brightness of Light Bulbs in Series and Parallel Circuits

In this video, we explain why the effect on brightness of light bulbs differs between series and parallel circuits (with calculations).

Power in Series vs Parallel Circuits

When analysing the brightness (power) of light bulbs, we need to consider whether they are connected in series or parallel. Assuming the total voltage (e.g. battery) remains constant, the different effects of series and parallel connections on power is best demonstrated by considering the voltage drop across each light bulb and by using the equation:

$$P = \frac{V^2}{R}$$

In a series circuit, the current through each bulb is the same. However, if the bulbs have different resistances, they will have different voltage drops across them and therefore will dissipate different amounts of power. This variance in power dissipation will result in different brightness levels:

Adding a light bulb in series will reduce the brightness of the light bulb(s) already in the series circuit. This is because an extra light bulb in series increases the total resistance, and decreases the total current. Since the current passing through each light bulb is now reduced, it becomes dimmer.

If the light bulbs are identical (equal resistance), they will each drop an equal fraction of the total voltage, and thus each will have the same brightness.

If the light bulbs have different resistances, the one with higher resistance will have a greater voltage drop, transform more electrical energy into light, and thus will be brighter.

In parallel circuits, each bulb has the same voltage across it (equal to the supply voltage) regardless of how many light bulbs there are.

Adding a light bulb in parallel will not affect the brightness of light bulb(s) already in the parallel circuit. This is because an extra light bulb in parallel decreases the total resistance, and increases the total current. This new increased total current is re-distributed across the light bulbs such that the voltage across each light bulb remains unchanged.

If light bulbs are identical (equal resistance), each bulb receives the full supply voltage and will have the same brightness, which will typically be brighter than if the same bulbs were in series (because they each get the full voltage)

If the light bulbs have different resistances, a bulb with lower resistance will draw more current and thus will be brighter than a higher-resistance bulb.

Calculation Example

The difference in power can also be understood by recognising that the total resistance of an electric circuit would differ if light bulbs are connected in series versus if they are connected in parallel. When identical light bulbs are connected in parallel, the total electrical resistance is lower than if they were connected in series.

Suppose we have a 2 `\Omega` resistor supplied by a 10 V battery as shown.

The power dissipated from the light bulb would be equal to the power delivered by the battery (conservation of energy).

As such, the power of the light bulb is given by:

$$P = \frac{V^2}{R}$$

$$P = \frac{(10)^2}{2} = 50 \text{ W}$$

Light Bulbs in Series

Now, let's consider a series circuit with two 2 `\Omega` light bulbs.

The total resistance would be

$$R_{\text{series}} = 2 + 2 = 4 \, \Omega$$

This increased resistance reduces the total current which is given by

$$I = \frac{V}{R} = \frac{10}{4} = 2.5 \text{ A}$$

The power dissipated from each light bulb in series circuit is given by

$$P = I^2R = (2.5)^2(2) = 12.5 \text{ W}$$

Light Bulbs in Parallel

If the light bulbs are connected in parallel, the new total resistance would be

$$\frac{1}{R_{\text{parallel}}} = \frac{1}{2} + \frac{1}{2}$$

$$R_{\text{parallel}} = 1 \, \Omega$$

Since the total resistance of the parallel circuit is lower and both circuits have the same voltage, the total current is greater in the parallel circuit (`V = IR`).

The power dissipated from each light bulb in parallel circuit is given by

$$P = \frac{V^2}{R} = \frac{(10)^2}{2} = 50 \text{ W}$$

Therefore, the brightness of light bulbs, when connected in parallel, remains unchanged.

Practical Demonstration

Example

A light bulb is connected to a battery in a series circuit. Explain the change in brightness of the light bulb if an identical light bulb is added to the circuit in series.

Solution:

Adding an identical light bulb in series doubles the total resistance of the circuit. This halves the current flowing through the light bulbs (`V = IR`). It also causes each light bulb to receive half as much energy from the battery; the voltage across each light bulb is halved.

Since electrical power is given by:

$$P = VI$$

$$P_{\text{new}} = (\frac{V}{2})(\frac{I}{2}) = \frac{P}{4}$$

The power dissipated in each light bulb is reduced by a factor of 4.