# Why is there a power drop in a serial circuit, but not in a parallel circuit?

I performed an experiment to compare the power dropped on a combination of a motor (small fan) and a lamp. I got the motor and a lamp from a lab, they are very old and all labels on the elements are gone. I connect them to a 3 V battery in both serial and parallel circuit as follows:

simulate this circuit – Schematic created using CircuitLab

In the top circuit, when the fan is spinning in full speed, I see that the lamp is very dim until I slow down the fan. In the bottom circuit, the brightness of the lamp will not be affected by the fan. I am trying to analyze this two situation in the following statements.

In the serial circuit, the voltage drop on motor and lamp add up to 3 V, but the current on both elements are the same. I am guessing the resistance of the lamp is much smaller than that of the motor so the voltage drop on the motor dominate. Most power is drop on the motor; I think that's the reason why the lamp is so dim. But what I don't understand is why the lamp becomes so bright if I hold the motor to stop the fan. Will stopping the fan cause the voltage drop on the motor?

In the parallel case, the voltage drop on lamp and motor are same. I know the current are different on elements. But in the case of using a battery as the power supplier, how is the current output determined?

I am thinking the total current output from the battery depends on the lamp and motor, in each path, the current is provided so to satisfy the power of the element. So it will try to output enough total current for each element to make the lamp bright and the fan spin fast. Is this correct?

• Redraw/simulate the circuit with the series resistance of the battery and it will all make sense. – winny Jul 17 '16 at 7:08
• @winny: He would see the same result with a regulated power supply which has a series resistance very close to zero. The effect s/he's seeing in the series circuit is due to the reduction in back EMF when the motor is slowed down. – Transistor Jul 17 '16 at 7:57
• @transistor Oh! Sorry. I overlooked that. – winny Jul 17 '16 at 8:49
• No problem. See my answer below. – Transistor Jul 17 '16 at 8:51

You are basically on the right track.

In both circuits, a perfect battery will output 3 V, and as much current as the load demands. Obviously a practical battery will have some internal resistance and a maximum current it can supply. The important point is that a perfect battery does not control the current - the load does.

In circuit 1, the effective resistance of the motor running at full speed is quite high. The motor consumes only the power needed to satisfy electrical and mechanical losses. Once you try to slow it down, it needs to provide mechanical output power, so it draws more current, and the effective resistance falls. So stalling the motor will make the lamp brighten.

In circuit 2, the lamp and motor each have 3 V across them, and the current drawn will be the sum of the lamp current and the motor current. So stalling the motor will cause a high current to be drawn, but if the battery is perfect, the lamp brightness will be unaffected.

A DC motor acts as a generator when the shaft is turned. The voltage at its terminals will be proportional to the speed.

simulate this circuit – Schematic created using CircuitLab

Figure 1. A model of a DC motor showing its internal series resistance and the back-EMF generated at various speeds.

The coil winding in a DC motor has some resistance so any time current flows there will be some losses given off as heat. This means that we coupled identical motors together, connected one to a battery and the other to the lamp, we would always get less energy out than we put in. The resistance also determines the maximum current that will flow into the motor.

In Figure 1a the motor is stopped. Therefore it is generating a back-voltage or back-EMF (electro-motive force) of 0 V. The only thing limiting the battery current is the 1 Ω internal resistor. The current will be given by $I = \frac {V}{R} = \frac {3}{1} = 3~A$.

In Figure 1b the motor is running at half speed. Therefore it is generating a back-voltage or back-EMF (electro-motive force) of about 1.5 V. Now there is 3 V at the top of the resistor and 1.5 V at the bottom giving 1.5 V across the resistor. The current will be given by $I = \frac {V}{R} = \frac {3 - 1.5}{1} = 1.5~A$.

In Figure 1c the motor is running as fast as it can. The limits will be caused by bearing friction, air resistance and losses in the windings. The back-EMF has now reached 2.5 V so the current drawn will be $I = \frac {V}{R} = \frac {3 - 2.5}{1} = 0.5~A$.

Now replace my ammeter with your lamp - a crude current indicator - and it should be clear what you are seeing. Note that an ammeter would have a very low resistance (< 0.1 Ω, say) so that it doesn't add too much resistance to the circuit whereas your lamp's resistance would be several ohms.

• Thanks. I read this statements several times and it makes sense to me.However, I still don't understand how to get the back-EMF from speed, is it from the conservation of energy, i.e. supplied electric energy should be used to compensate the motor's kinetic energy (if ignore heat lost)? – user1285419 Jul 17 '16 at 17:37