Q:

I'm still confused about bulbs in series having same amount of current flowing through them When current passes through bulb A from battery doesn't the resistance of bulbA cause a decrease in this current when it leaves bulb A and goes to bulb B so current entering B is less than current that entered A ? What's wrong with my thinking?

- Dolores Vermont (age 72)

Chesterfield

- Dolores Vermont (age 72)

Chesterfield

A:

Hi Dolores,

Remember that current is the flow of electrons. If one Coulomb of electrons goes past a certain point every second, then we say the current at that point is 1 Ampere. For your situation, let's imagine the circuit has one battery connected in series with two bulbs, A and B. In direct current, which is the current you have with a battery (as opposed to alternating current with a generator), charges leave the battery and travel around the circuit. Since we only have one wire (with two bulbs), once the charges leave the battery they are constrained to only travel in this wire. They go into the first bulb, A, but as many charges go out as go in. If this weren't true, then charges must build up at the light bulb, since many charges are entering the bulb and only few are leaving. You can imagine the charges moving through this wire to be like water moving through a pipe: the amount of water that goes into any section of pipe is equal to the amount of water leaving that section. Just as water can't build up in some segment of the pipe, electrons can't build up in some segment of the wire!

Since the current must leave, then it next reaches the second light bulb, where the same story is true- the current is the same on both sides of the bulb. In fact, the current can only change when there is a junction, and even then the total amount of current entering a junction is equal to the total amount of current leaving a junction.

You are right that something changes in your circuit across the light bulb though---the voltage on either side of the light bulb is not the same. There is a specific voltage drop at your bulb (equal to IR, since your bulb acts as a resistor, dissipating energy), so the voltage before your light bulb is I*R larger than the voltage at a point right after the bulb. Also, if you have two bulbs in series, you will notice that, upon attaching the second, the bulbs get dimmer. Now, compared to before, they are getting less current (since the net resistance of the system increased), but each light bulb is receiving exactly the same current, as the electrons travel from the battery, through each bulb, and then back to the battery. They do have a smaller voltage drop than either bulb would have in a circuit by itself, which is why they appear dimmer (since brightness is a function of the power).

Hope this is clear! If not, feel free to follow up.

Ben M.

Remember that current is the flow of electrons. If one Coulomb of electrons goes past a certain point every second, then we say the current at that point is 1 Ampere. For your situation, let's imagine the circuit has one battery connected in series with two bulbs, A and B. In direct current, which is the current you have with a battery (as opposed to alternating current with a generator), charges leave the battery and travel around the circuit. Since we only have one wire (with two bulbs), once the charges leave the battery they are constrained to only travel in this wire. They go into the first bulb, A, but as many charges go out as go in. If this weren't true, then charges must build up at the light bulb, since many charges are entering the bulb and only few are leaving. You can imagine the charges moving through this wire to be like water moving through a pipe: the amount of water that goes into any section of pipe is equal to the amount of water leaving that section. Just as water can't build up in some segment of the pipe, electrons can't build up in some segment of the wire!

Since the current must leave, then it next reaches the second light bulb, where the same story is true- the current is the same on both sides of the bulb. In fact, the current can only change when there is a junction, and even then the total amount of current entering a junction is equal to the total amount of current leaving a junction.

You are right that something changes in your circuit across the light bulb though---the voltage on either side of the light bulb is not the same. There is a specific voltage drop at your bulb (equal to IR, since your bulb acts as a resistor, dissipating energy), so the voltage before your light bulb is I*R larger than the voltage at a point right after the bulb. Also, if you have two bulbs in series, you will notice that, upon attaching the second, the bulbs get dimmer. Now, compared to before, they are getting less current (since the net resistance of the system increased), but each light bulb is receiving exactly the same current, as the electrons travel from the battery, through each bulb, and then back to the battery. They do have a smaller voltage drop than either bulb would have in a circuit by itself, which is why they appear dimmer (since brightness is a function of the power).

Hope this is clear! If not, feel free to follow up.

Ben M.

*(published on 03/28/2011)*

Q:

A 230v,100w bulb and a 230v,25w are connected in series across 230v DC supply. What will be the power consumed by the circuit. How to find the resistance and current and power consumed by it. If the current decreases in this circuit can we find it. but while calculating we will assume the current as same. I am not clear with the current differs.. Pls expalin.

- Krishna (age 21)

Madurai,Tamilnadu,Indi

- Krishna (age 21)

Madurai,Tamilnadu,Indi

A:

Each bulb has an effective resistance R = V^{2}/W. Connecting two bulbs in series gives a total resistance of R_{tot} = V^{2}*(1/W_{1} + 1/W_{2}). The current and hence the total power is decreased and the total light output is diminished. When I put in the numbers I got 20 Watts total.

There is a small correction due to the fact that the voltage across each bulb is slightly decreased. Since the filament resistance drops slightly with voltage the actual resistance is not the same. In this case it is a rather small effect.

LeeH

There is a small correction due to the fact that the voltage across each bulb is slightly decreased. Since the filament resistance drops slightly with voltage the actual resistance is not the same. In this case it is a rather small effect.

LeeH

*(published on 06/17/2011)*