Ohm's law states that voltage is resistance times current. In a resistor circuit, knowing two of voltage, current, or resistance, you can calculate the third.
Actually, this applies to any circuit, be it resistor, capacitor, or inductor. Ohm's law still applies - it just gets more complex when the phase angle of current is not the same as the phase angle of voltage.
If you are looking for the resistance of each resistor in either a series circuit or a parallel circuit you must measure the current I and the voltage V for each resistor. Then calculate its resistance using Ohms Law R = V / I where I = current (Amps), V = voltage (Volts) and R= resistance (Ohms).
Resistance (Ohms) = Potential Difference (Volts) / Current (Amps) So, 12/0.25 = 48 Ohms.
This is a direct consequence of Ohms Law. Since each new resistor connected in parallel will allow more current to flow, the resistance of the circuit must be lower. R = E/I. Since I (current) has increased, and the voltage E is still the same, it follows that R (resistance) must be smaller. That's the way Ohms Law works.
Using Ohms Law, the answer is 120/0.5 = 240 Ohms.
A resistor is a resistor. Plain and simple. By Ohm's Law, resistance in ohms is voltage in volts divided by current in amperes. The difference lies in application, not in the resistor itself. A normal resistor will introduce a voltage drop or current that makes some effect in the circuit, based on some design criteria. A bleeder resistor, on the other hand does not really affect the circuit - it is only there to "bleed off", or discharge, capacitors when the power is turned off. Consequently, a bleeder resistor will typically have a higher resistance than a normal resistor but, again, the issue is circuit design, not the resistor itself.
Depends on the device. If it is a resistor and you have a fixed voltage then the circuit will obey Ohms law. Voltage = Current x Resistance. So if R increases by adding more resistors in series and the voltage is constant, the current will decrease.
You describe a series circuit. This answer is for a series circuit. The total impedance is 16 Ohms, 12 + 4. The current by Ohm's Law (Amps = Volts / Ohms) is 40 / 16, or 2.5 Amps. Kirchoff's Current Law says that the current at every point in a series circuit is the same, so the current flowing into the 1st device, the 12 Ohm resistor, is 2.5 Amps.
9V by using ohms law
No one is going to be able to tell you that. You are looking for the measured voltage, so go and measure it. In any case, if you were just looking for the voltage it will depend on the circuit current. You can work it out using ohms law (Voltage = Current * Resistance).
The letter R is used to represent resistance. For instance, the R in a circuit is said to be 52 ohms. Just that simple.
R2 = 3 ohms Explanation: For a circuit you can use ohms law where: V=I*R Where V is the voltage difference throughout the surface, I is the current, and R is the total resistance of the circuit. In your case you want to find the resistance so you have to change the formula to: R=V/I R of first circuit = 25volts/12.5amps = 2 ohms R of second circuit= 25 volts / 5 amps = 5 ohms The resistors here are connected in series which means that the resistance of the two can be added together. This gives you: Rtot= R1+R2 we found R of the first resistor by calculating the resistance in the first circuit. We also found Rtot which is the resistance in the second circuit, when you connect the two resistors together in series. Rtot=2 ohms+R2 5ohm=2ohms+R2 R2 = 3 ohms If the resistors where connected in parallel you cannot simply add the resistance. In that case: (1/Rtot)=(1/R1)+(1/R2) Hope that helps
Just use Ohm's Law Voltage = Current x Resistance Amps = Voltage Divided By Resistance Amps = 120 / 260