If I have a device that requests 6V input and I have an a/c adapter that outputs 9V, will the device draw only the 6V it needs or will it try to take all 9V and burn itself out? In this case the circuit is simply a light. It has the switch and the bulb with an optio of being powered by 4 AA batteries or 6V input. No amp requirement specified.
Watt’s it matter?
Quick lesson in electrical theory. Electricity has two components, voltage and current. They are related by a characteristic of the circuit called resistance according to Ohm’s law,
voltage=current *resistance
Mathematically, current=voltage/resistance. If you increase voltage, current will go up.
If you think of electricity like a bullet, voltage is the gunpowder; it provides the energy. Current is the slug; it does the work.
Or you can look at it like a water pipe. Voltage is like water pressure, where current is like water flow. The higher the pressure, or voltage, the more water or current will flow.
If the circuit calls for 6V and you attach it to 9V, the circuit will draw 150% of design current because the input voltage is 150% of design.
Current is what causes things to light up, heat up and burn out. In our simple circuit, the additional voltage will cause the light to burn brighter, and to burn out much quicker.
In other applications, say computer circuitry, proper input voltage is essential for correct operation.
Excellent! Thank you. I used to know some of this but it seems that my memory regarding electricity has burned out. I think I’ll go to
Radio ShackSpark Electronics and buy a basic kit to teachmeone of the children basic electronics.