Ohm's Law: Current is voltage divided by resistance.
12 volts divided by 470 ohms is 25.5 milliamperes.
Power is voltage times current, so power in this case would be 12 volts times 25.5 milliamperes, or 0.306 watts. As a result, you need at least a one half watt resistor, and I would prefer a one watt resistor, because it is going to get a bit warm, and a margin of safety is always a good thing. This is particularly true when you consider tolerances, such as the battery actually putting out 13 or 14 volts and the resistor being on the low side, at 470 - 10%, or 423 ohms. (0.463 watts - too close to one half watt for comfort)
A circuit with a 2 ohm resistor and a 4 ohm resistor in series with a 12 volt battery will have 2 amps flowing through each resistor. The current is the same in each resistor because they are in series, and a series circuit has constant current throughout.
12 milliamps
yes... this is possible if a diode i connected in reverse bias with a battery and a resistor for example. A diode in reverse bias means its anode will be connected to positive terminal of the battery and its cathode to the negative terminal of the battery. In such a case, minimal current flows through the circuit which can be neglected.
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.
Questions seem to be two! First is the direct current in a circuit with 12 V battery and 3 ohm resistor in it. The answer is 4 ampere if there is negligible internal resistance in the battery. By ohm's law V = R I So, I = V/R Second one is the alternating current with 120V given to 60 W bulb. The answer is 0.5 ampere. This 0.5 is the rms value of alternating current. Power P = V I So, I = P/V
A circuit with a 2 ohm resistor and a 4 ohm resistor in series with a 12 volt battery will have 2 amps flowing through each resistor. The current is the same in each resistor because they are in series, and a series circuit has constant current throughout.
To find the current through the 40 ohm resistor, first calculate the total resistance of the parallel circuit: 1/Rt = 1/120 + 1/60 + 1/40. Then, calculate the total current using Ohm's Law, I = V/Rt. Finally, use the current divider rule to find the current passing through the 40 ohm resistor.
Current flows through a resistor, not across it.
When an electric current flows through a resistor, the resistor resists the flow of the current, causing a decrease in the current. This decrease in current is proportional to the resistance of the resistor, as described by Ohm's Law.
12 milliamps
Normally through the resistor's internal construction. It flows through any part of the resistor that has low resistance- be it anywere. And then there's this. It might be that one should consider that current flows through a resistor and voltage is dropped across a resistor. Perhaps this is where the question began. The former is fairly straight forward. The latter can be vexing. Voltage is said to be dropped across a resistor when current is flowing through it. The voltage drop may be also considered as the voltage measureable across that resistor or the voltage "felt" by that resistor. It's as if that resistor was in a circuit by itself and hooked up to a battery of that equivalent voltage.
All the way along when the crocodile clip is connected to a resistor, when the other end of the resistor is connected to the other side of the battery.
0.81 APEX
p=I*I*R ,P=V*V/R;where I is the current passing through the resistor, and V is the voltage across resistor, and R is the Resistance of the resistor,
Voltage = Current * ResistanceVoltage = 12VResistance = 10 ohmsCurrent = Voltage/ResistanceCurrent = 12V/10 ohmsCurrent = 1.2 Amps
What is the current running through resistor four?1 amps..!What is the current running through resistor one? 3 amps...!What is the current running through resistor three? 2amps..!What is the current running through resistor five? 3 amps..!What is the voltage drop running through resistor five? 45 volts...!What is the equivalent resistance through the parallel portion of the circuit? 6 ohmsAnswerA resistor is a conductor, albeit one with a higher resistance than a length of wire, so current passes through it without any problem. The magnitude of the current will, of course, be somewhat lower because of the additional resistance.
If they're in parallel, then each resistor acts as if it were the only one,and the presence of any others is irrelevant.The current through the 60-ohm resistor is I = E/R = (120/60) = 2 amperes.