You can only use a resistor to drop a voltage at a constant current. If you know the current, use Ohm's law to calculate the resistor value.
you better not. you need a 9V regulator for this, not a resistor!
No, of course not. The 6v adapter only produces 6v, while the load expects 9v.
Substituting a 9V DC power supply for a 6V DC supply is generally not advisable, as it can lead to overvoltage conditions that may damage the device designed for 6V operation. Components may overheat or fail, and the device may not function correctly. Always check the specifications of the device to ensure compatibility with the power supply voltage. If a 9V supply is required, consider using a voltage regulator or a suitable step-down converter to safely reduce the voltage to 6V.
9v
Even though it is connected to a 9 volt source, it is still a resistor.
If a 9V supply is connected to a 100-ohm resistor, then the current is not 2 Amps,or even close to it.I = E/R = 9/100 = 0.09 amp.
Yes, connecting a device designed for 6V to a 9V power supply can potentially damage the device. The excess voltage may cause overheating, overcurrent, or even permanent failure of the components. Always check the voltage specifications of your device to ensure safe operation. If in doubt, using the correct voltage is crucial to avoid damaging your equipment.
No, a 9V 1A power supply cannot replace a 6V 300mA power cord. The voltage of the replacement power supply is higher than the original required voltage, which could damage the device. Additionally, the current rating of the replacement is sufficient, but it’s essential to use a power supply that matches both the voltage and current specifications of the device. Always adhere to the manufacturer's requirements for voltage and current to ensure safe and proper operation.
No. The adaptor will overheat.
0.81 APEX
Yes, the manufacturer specifies what the working voltage of a device will be and that voltage has to be adhered to.
In order to determine what size of resistor is required to operate an LED from a 9V battery, first start by knowing the current and voltage required for the LED. That information is available in the LED's specifications. For discussion purposes, lets assume a typical LED at 2.5V and 50mW. The translates to a forward current of 20mA. Build a simple series circuit containing a 9V battery, a resistor of an as yet unknown value, and the LED. By Kirchoff's current law, the current in the LED is the same as the current in the resistor, which is also the same as the current in the battery. This is 20ma. By Kirchoff's voltage law, the voltage across the LED plus the voltage across the resistor equals the voltage across the battery. This is 6.5V. (9 - 2.5) By Ohm's law, resistance is voltage divided by current, so the resistor is 6.5 / 0.02, or 325 Ohms. The nearest standard value to that is 330 Ohms. Cross check the power through the resistor. Power is voltage times current, or 6.5V times 0.02A, or 0.13W. A half watt resistor is more than adequate for this job.