23 volts across 470 ohms will dissipate about 1.1 watts.
Power equals voltage squared divided by resistance.
To calculate the resistance in ohms, you also need to know the voltage. The formula to find the resistance in ohms is R = V^2 / P, where R is the resistance in ohms, V is the voltage, and P is the power in watts.
97 k ohms is equivalent to 97,000 ohms. Just multiply the value in kiloohms by 1,000 to convert it to ohms.
It depends on the load's resistance (ohms). By using the formula P (watts) = V^2 (volts) / R (ohms), and assuming a load resistance of 1 ohm, the voltage would be 1 volt for 1 watt.
To find the resistance of a 100W bulb operating at 200V, you can use the formula ( R = \frac{V^2}{P} ), where ( R ) is resistance, ( V ) is voltage, and ( P ) is power. Substituting the values, ( R = \frac{200^2}{100} = \frac{40000}{100} = 400 ) ohms. Therefore, the resistance of the 100W bulb at 200V is 400 ohms.
470 mL = 0.47 LTo convert from mL to L, divide by 1000.
To calculate the resistance in ohms, you also need to know the voltage. The formula to find the resistance in ohms is R = V^2 / P, where R is the resistance in ohms, V is the voltage, and P is the power in watts.
470.... apparently :P
250 watts approximatelyI assume you have a constant voltage supply. According to P=V^2/R => P*R=V^2, you have a 44.7 volt supply. If you change Resistance to 8 ohms, P=44.7^2/8 => P=250 watts.
The formula you are looking for is, R = Volts (squared)/Watts.
Though it is tempting to say the difference is 2 ohms (8 ohms minus 2 ohms equals 6 ohms), lets look at some things. The 6 ohms is 3/4ths the resistance of the 8 ohms. If the resistances are loads, the 6 ohm load will draw 1/3rd more current than the 8 ohm load. The 8 ohm load will draw 3/4ths as much as the 6 ohm load. Those are some differences between 6 ohms and 8 ohms.
Watts = Volts x Amps x Powerfactor Volts = Amps x Ohms Note: Power Factor is one for resistance loads and decreases toward zero for loads like motors. watts=VI=v*v/R=I*I*R volt=IR=square root of PR=P/I Ampere=V/R=P/V=square root of P/R ohms=V/I=V*V/P=P/I*I
Earthing of the negative terminal is a common practice and will not affect the potential across the resistors in the circuit. Calculating this is simple Ohm's Law. For a series circuit: R(total) = R1 + R2 + R3....... Rn In this case: Rt = 120 + 470 = 590 ohms Per ohms law E = I x R. We need to find the current flowing throught the circuit so: I = E/R = 12 / 590 = .0203 amps or 20.3 milliamps The potential across an individual resistor is calculated again by E = I x R So for R1 (120 ohms): E1 = I x R1 = .0203 x 120 = 2.44 V So for R2 (470 ohms): E2 = I x R2 = .0203 x 470 = 9.54 V For a check E = E1 + E2 = 2.44 + 9.54 = 11.98 The .02 difference is due to the use of significant figures.Another AnswerEarthing the negative terminal will have no affect on the potential DIFFERENCE across each resistor. 'Potential' exists at a single point. 'Potential difference' exists between two points.
It is 470.
81% of 470= 81% * 470= 0.81 * 470= 380.7
30% of 470= 30% * 470= 0.3 * 470= 141
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)
0.069444444444444444444444444444444444 ohms. P/E^2=R. P = power in watts. E = electricity in volts. R = resistance in ohms.