# Series and parallel circuits

• Engineering
Jack John
if we have 2 circuits. series and parallel. both of them have 3 light bulbs connected to them. 1 ohm 2ohms 3ohms. both circuits have these light bulbs. will the 3ohm light brighter in both circuits?why or why not? I am thinking yes because the higher the resistance, the more collisions there will be so more electrical energy is converted into heat and light, but I am not sure, so willl the 3ohm light brighter in both circuits? why or why not .

Krasner
No. The Three Ohm light bulb will be the brightest only if it is connected parallel to the other light bulbs. This is because the voltage drop on this light bulb will be equal to the voltage of the battery. However, in series the voltage drop on this light bulb will be smaller than the initial voltage. Thus in a series, the 3 ohm light bulb might actually be the dimmest one. A parallel circuit will guarantee it to be the brightest light bulb.

Homework Helper
Gold Member
No. The Three Ohm light bulb will be the brightest only if it is connected parallel to the other light bulbs. This is because the voltage drop on this light bulb will be equal to the voltage of the battery. However, in series the voltage drop on this light bulb will be smaller than the initial voltage. Thus in a series, the 3 ohm light bulb might actually be the dimmest one. A parallel circuit will guarantee it to be the brightest light bulb.

It's actually the other way around.

In series the current is the same in all three so the best equation to use for the power is P = R I^2 which shows that the largest resistance will dissipate the most power.

In parallel they all have the same voltage so it's better to use P = V^2/R which shows that the largest resistance will dissipate the smallest power.

rolerbe
Call the bulbs B1, B2 and B3.

In series, the total resistance is 6 ohms. So the current is V/6. The power in each bulb is the same: PB1,PB2,PB3 = 1/6 V^2, so they are equally bright. This is because each drops a different amount of voltage. VB1 = V/6, VB2 = V/3, VB3 = V/2.

In parallel, the voltage drop for each bulb is the same: V. The current is different. IB1 = V/1, IB2 = V/2, IB3 = V/3. So the power in each bulb is PB1 = V^2, PB2 = 1/2 V^2, PB3 = 1/3 V^2.

So, in series all the bulbs are the same brightness. In parallel, B3 is the dimmest of the 3. But its still brighter than in series.

rolerbe
Corrected: Call the bulbs B1, B2 and B3.

In series, the total resistance is 6 ohms. So the current is V/6. CORRECTION: each bulb drops a different amount of voltage. VB1 = V/6, VB2 = V/3, VB3 = V/2. So, the power in each bulb is PB1 = 1/36*V^2, PB2 = 1/18*V^2, PB3 = 1/12*V^2.

In parallel, the voltage drop for each bulb is the same: V. The current is different. IB1 = V/1, IB2 = V/2, IB3 = V/3. So the power in each bulb is PB1 = V^2, PB2 = 1/2 V^2, PB3 = 1/3 V^2.

So, in series B3 is brightest. In parallel, B3 is the dimmest of the 3. But its still brighter than in series.

saiedali2005
series the current is constant ,parallel he voltage is constant