Question is not clear.
Is question asking about the battery's internal resistance ?
AnswerResistance is not affected by voltage. The resistance of a material depends upon the length, cross-sectional area, and resistivity of that material. As resistivity is affected by temperature, resistance is also indirectly affected by temperature.
That is 366,300 ohms.
It is simple LEDS do need a definite voltage drop to work must have a minimum current to operate. for a 9v battery source and a LED of 3.2v operating at 20ma the solution will be 9v-3.2=5.8v therefore to limit the current to 20ma a resistor of 5.8v/0.02=290 ohms not such an animal so a 270-300 ohms will work. Final note make sure that the LED power is not exceeded 3.2v x0.02ma=63 mw
8x12=96
One tera means 1012. So 1 Teraohm = 1012 ohms. One trillion ohms
R = U / I = 240 / 20 = 12 Ohms.
Current I = V/R V = 9V R = 100ohm I = 9V/100ohm = 90mA
9V by using ohms law
Current I = V/R V = 9V R = 100ohm I = 9V/100ohm = 90mA
A: No problem but don't expect the power to increase
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, a 9v 1100mA power adapter work with your 9v 500mA device. The mA number is the maximum amount of amperage that the adapter will produce without overloading itself. At 500 mA the adapter will only be working at half load.
total voltage = 4.5V, total resistance = 3.5 ohms, loop current = 4.5V / 3.5 ohms = 1.286Atotal voltage = 9V, total resistance = 4 ohms, loop current = 9V / 4 ohms = 2.25Atotal voltage = 13.5V, total resistance = 4.5 ohms, loop current = 13.5V / 4.5 ohms = 3Aetc.There is no solution to your problem conditions.
For an end-result of 10 ohms, you can have 1 resistance of 10 ohms, or 2 parallel resistance of 20 ohms each, or 3 resistances of 30 ohms each, etc.
Current, I is equal to V (voltage) divided by R (resistance); Hence: I=V/R = 9V/200 ohms: I = 0.045 Amps, or 45ma (milliamps).
1000 ohms.
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.