Basic DC Circuit Question

The motor/fan and light bulb have a low resistance and use a high current. The piezo buzzer has a very high resistance and uses a very low current that prevents the high current devices to work in series with it. The motor/fan and light bulb pass plenty of current to the buzzer.

Light, buzzer and fan are loads for the system. The two batteries are the power source for the system. The two batteries are connected in series (hence the voltage is added: 1.5+1.5=3V) and the current is perhaps 1-3A (because they are big batteries).

The loads are classified by their voltage and current requirements (you can also specify the voltage or power) and when you connect two loads in series, they divided the applied voltage (3V in this case) between them. it is not going to be shared equally. Consider the lamp and the fan in series. It is very likely that both will work but with reduced capacity. The light will be dim and the fan will be slow. The light may be very dim, perhaps. Why?

We need to have some idea about how the voltage and current are shared between devices. When you connect them in series, the same current flows through all of them. The buzzer can work at a low current and hence the buzzer works. The fan and light pass the current but because it is low, they do not seen to work.

when you have the fan and the light in series, the fan can work at a lower current and hence only the fan works. Because they are in series, the same current goes through the lamp and the fan but the lamp glows only very dim because the current is low.

Hi,

A maybe pedantic clarification:

c_mitra said:

and the current is perhaps 1-3A (because they are big batteries).

Click to expand...

This may sound as if the batteries determine the current.
But usually the current is determined by the load.

No connected load = no current.

****
While you may use a DVM to measure the voltage of a battery (or other source)
.. you should not try to measure the battery current directly.

Usually you measure the current by connecting: battery - amperemeter - load - (back to battery).
But then the current is determined by the load. You may call it "load current", although the current is the same in every point of the loop (Battery, wires, load, amperemeter...)

****
Exceptions (rare) are sources with high internal impedance.
Like current sources, current loops (for industrial sensor measurement, for example) etc.

Klaus

KlausST said:

Hi,

A maybe pedantic clarification:

This may sound as if the batteries determine the current.
But usually the current is determined by the load.

No connected load = no current.

Klaus

Click to expand...

Sorry that I was not clear enough.

When I said that the battery will have a voltage of about 3V, I meant the open circuit voltage (no load connected).

Similarly when I say that the batteries may deliver about 1-3A current, I meant the short circuit current (a load with zero resistance is connected).

Without any load connected, the battery will have the highest voltage (open circuit)
With a load of zero resistance connected, the battery will have highest current (short circuit current)

Why these are relevant? Because all real batteries have some finite internal resistance. Ideal batteries have zero internal resistance (ideal voltage source) and their potential does not change when you connect with the load. When shorted, they can supply infinite current. Unfortunately ideal voltage sources do not exit.

The internal resistance of a battery becomes important when you consider the real power delivered by the battery to an external load. Because of the internal resistance, all batteries cannot deliver arbitrary power to an external load. Max power is delivered to an external load when the load resistance is equal to the internal resistance of the battery.

I am very sorry that I was not explicit enough.

The current into a load is determined by the load.

The max current that the battery can deliver is determined by the internal resistance of the battery.

The actual current a pair of D-cells can deliver into a load of zero resistance will be perhaps around 1-3A but this number is a guess. It will vary with the brand and will continue to drop during its useful life.

When you short the battery terminal, the terminal voltage drops to zero and the current becomes large.

When the load is absent, the terminal voltage reaches the max value but the current is now zero,

I hope I am bit more clear now.

Thanks everyone for taking the time to respond. All of this information is very helpful. I have one more question. Is there a possible voltage that could be applied and all three devices would work in series or will it always only be the buzzer that works for the reasons already explained above? I did connect it up to two 9-volt batteries in series (18 volts) and it was still only the buzzer that worked. Again, thanks for your responses.

Hi,

in a series connection of three different loads..
the voltage ratio is determined by the load resistance.

an example:
* load1 has 700 Ohms, Load 2 has 1000 Ohms, load 3 has 300 Ohms.

if you connect them in series you get:
700 Ohms ---- 1000 Ohms ---- 300 Ohms
resulting in a total resistance of 2000 Ohms

Since the voltage is proportional with the resistance...
2000 Ohms stands for 100%
So ...
* across 2000 Ohms there is 100% of the (battery) voltage
* across 700 Ohms there is 700/2000 = 35% of the voltage
* across 1000 Ohms there is 1000/2000 = 50% of the voltage
* across 300 Ohms there is 300/2000 = 15% of the voltage

****
But this really is basic electronics stuff.
Many others created tutorials and even tutorial videos about this.

I think it´s a good idea for you to read/view a couple of these tutorials.

.. a froum can not replace school, or reading books or tutorials

Klaus

The motor and light bulb draw and need a high current of maybe 500mA. But the piezo beeper draws and passes only a low current of maybe 10mA.
If the motor and light bulb barely work at 1/4 their rated current then to pass the calculated125mA, the beeper would need 125/10 x 3V= 37.5V which would burn it out.

Little 9V batteries have trouble producing more than about 50mA.

Rudy S said:

Is there a possible voltage that could be applied and all three devices would work in series or will it always only be the buzzer that works

Click to expand...

The short answer NO!

The long answer: The three devices have been designed to work with different currents. When the voltage ratings are close, you must connect the devices in parallel so that all of them get the same voltage.

Same way, if their current requirements are similar, you can connect them in series. In a series connection, the same current flows through all the components in the series connection.

If you look up a LED (datasheet), you will see that it has been specified with a minimum voltage and a specified current. That means it need a minimum voltage to turn on and then you must maintain a specified current. You can also force the specific current through the device and see that the voltage across the LED is just slightly above the minimum voltage. This is because a LED is typical non-linear device.

You can connect three lamps in series or in parallel. Why? because they can work at the same current and also at the same voltage. When in series, you need to use twice the voltage you used for each lamp. When used in parallel, your battery must supply twice the current. But that will surely not work if the two lamps are not rated for the same current or same voltage.

I hope I am clear.

If you are going to teach students about simple circuits then, at the least, you need to understand Ohm's law and how parallel and series circuits work.
Since you don't seem to have that understanding, I suggest you read some tutorials on that.