No. This would mean more power is used than is provided - an impossibility. It is possible that you may not be accounting for a power source in your circuit, or have the actual source modelled incorrecty.
In a DC circuit, the power dissipated by a resistance is (voltage across it)2 divided by 'R'.P = E2/R = (14.1)2 / 142 = 198.81/142 = 1.4 watts(rounded)
A resistor.
The power dissipated by a 10 ohm resistor with 800v across it is 64 kw.Ohm's law: current is voltage divided by resistancePower law: power is voltage times current, so power is voltage squared divided by resistanceDon't even think about trying this. 64 kw is a lot of power. The resistor will probably explode, or catch fire. At best, the 80 amps required will trip your circuit breaker, if you are lucky.
As you add more bulbs to a series circuit that means that the bulbs are in series to one another, therefore the total resistance is the sum of the individual resistance of the bulbs. If you add bulbs of the same resistance,then the rate at which the resistance changes will increase in a constant manner provided the current source is not altered. For instance if the bulb you are using is rated 20v,60w, then the current passing via the bulbs in series is the square of the voltage divided by the power in this case the current is approximately 7amperes.
A series dropping resistor is a resistor that limits the amount of current flow in a circuit.
The energy dissipated by a resistor in an RC circuit is calculated using the formula: Energy 0.5 C V2, where C is the capacitance of the circuit and V is the voltage across the resistor.
Depends on the current. Put a resistor in-line with the current, then measure the voltage across the resistor. V=RI. So, divide the measured voltage by resistor value. Be careful with the size of the resistor, as Power dissipated in a resistor is R*I^2 or V^2/2. So, a 1-Amp current into a 1 Ohm resistor will result in a 1Watt power dissipated in the resistor. If it's too small, it'll burn. Also, notice that if you do that, you haven't measured the current in the original circuit. You've measured the current when an extra resistor is installed in the original circuit, and that's different.
what does a resistor in an ampliflyer circuit
if not disconnected you will measure the resistance of the circuit in parallel with the resistor.
In a DC circuit, the power dissipated by a resistance is (voltage across it)2 divided by 'R'.P = E2/R = (14.1)2 / 142 = 198.81/142 = 1.4 watts(rounded)
AS:total power = P1 + P2 + P3 + .......so,total power = 5(50mW)= 0.25Wtotal power dissipated by the five resistors is 0.25W.
The power dissipated in a voltage divider circuit is given by the formula P = V^2/R, where V is the voltage across the resistor and R is the resistance of the resistor. If the resistance in the voltage divider circuit is increased, the power dissipated in the circuit will decrease. This is because as resistance increases, the current flowing through the circuit decreases, leading to less power being dissipated as heat in the resistors.
A resistor.
Any part of a circuit that has a voltage drop across it is a resistor.
The power dissipated by a 10 ohm resistor with 800v across it is 64 kw.Ohm's law: current is voltage divided by resistancePower law: power is voltage times current, so power is voltage squared divided by resistanceDon't even think about trying this. 64 kw is a lot of power. The resistor will probably explode, or catch fire. At best, the 80 amps required will trip your circuit breaker, if you are lucky.
It is used to vary the voltage/current flow in a circuit.
The power dissipated by the complete circuit, no matter whether it's a series or parallel one, is the simple sum of the power dissipated by each component of the circuit.