I think that the anwer would be 40 ohm's
circut of an amplifier
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
Series - parallel circut
It is called the loopdie loop circut.
when the switch the emitter-base junection is an opnd circut and heance the value of input or base curent is zero.
circut
It heats up by opposing the flow of charge through the circui- Apex
current will increase
the amount of power going to the object will will cause the circut breaker to blow. if the object is not connected to a breaker then the object will catch on fire.
Parallel Circut
Ohm's law states that V=IR, that is, Voltage is equal to the current through the circuit times the resistance.Manipulating this equation gives I=V/R, so I=60/12 = 5 A
A voltmeter measures the voltage present in the circut. Voltage is the amount of energy available to push charges through an electric circut.
fuse
The current /electricity will originate in a battery or a dynamo.
move the magnet faster
The current passing through them - since all wires have resistance.
your mom is so fat she cloggs up all the paths for the current