half its output value
.9 watts.
The resonant frequency is set by the L and the C: 900 microhenrys and 0.014 microfarad would resonate at a frequency given by: F = 1/ [2 pi sqrt(LC)] If the components are in 'micros' the answer is in Megahertz. In this case the resonant frequency in MHz is: F = 1/ [2 pi sqrt(900 x 0.014)] or 0.0448 MHz, 44.8 kHz. The reactance of the inductor and the capacitor is 253 ohms, so adding a 1000 ohm resistor in parallel would give a tuned circuit with a Q of 1000 / 253 or 3.9.
A circuit has an applied voltage of 100 volts and a resistance of 1000 ohms. The current flow in the circuit is 100v/1000ohms which would equal .1.
If they're connected in series the total resistance is 2000 ohms. If they're connected in parallel the resistance is 500 ohms.
The capacitive reactance is approximately 4 kΩ .
The power will be the product of the square of the current and the resistance of the load. The fact that the circuit is a parallel circuit is irrelevant to this question.P = I2R = 0.032 x 1000 =0.9 W
.9 watts.
I don't know what the parallel circuit has to do with it. You've onlygiven me a resistor and the current through it.When 0.03A of current passes through a 1,000Ω resistor, the resistordissipates energy at the rate of 0.9 watt.
The resonant frequency is set by the L and the C: 900 microhenrys and 0.014 microfarad would resonate at a frequency given by: F = 1/ [2 pi sqrt(LC)] If the components are in 'micros' the answer is in Megahertz. In this case the resonant frequency in MHz is: F = 1/ [2 pi sqrt(900 x 0.014)] or 0.0448 MHz, 44.8 kHz. The reactance of the inductor and the capacitor is 253 ohms, so adding a 1000 ohm resistor in parallel would give a tuned circuit with a Q of 1000 / 253 or 3.9.
A circuit has an applied voltage of 100 volts and a resistance of 1000 ohms. The current flow in the circuit is 100v/1000ohms which would equal .1.
Yes. The equivalent resistance of resistors in parallel is written as 1/Req=1/R1+1/R2+1/R3+... which, in this case, would be 1/Req=1/1000+1/1000+1/1000+1/1000=0.004. This means that Req=1/0.004=250Ohms.
circuit court
2003 Ohms. R = (R1 x R2)/( R1 + R2) Where R = 667 and R1 = 1000 then R2 = 2003
The total wattage of those 3 devices is 1860 Watts. Divide the Watts by the 120 Volts in the circuit and you get 15.5 Amps. That exceeds the 15 Amp fuse.
The Bathurst 1000 race is usually contested over a distance of 1000 kilometers, which is equivalent to approximately 161 laps around the Mount Panorama Circuit.
A parallel run of 750 MCM AWG conductors will handle 1000 amps. if we want 1000amps to flow, 250sqmm cable is enough.
For a 1000 amp circuit, you would typically need a very large wire gauge, such as 3/0 or 4/0 AWG copper wires to safely handle the high current without overheating. It's always recommended to consult with a professional electrician or refer to the National Electrical Code (NEC) for specific wire sizing requirements for high-amperage circuits.