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The time constant for a RC circuit is given by?
In an RC circuit the time constant is found by R x C. T = R x C to be precise.It is the time required to charge the capacitor through the resistor, to 63.2 (≈ 63) percent of full charge; or to discharge it to 36.8 (≈ 37) percent of its initial voltage. These values are derived from the mathematical constant e, specifically 1 − e − 1 and e − 1 respectively.
Time constant of LC circuit is?
T=sqrtLC
What is third harmonics in ac circuits?
It depends on the frequency in the expression of the function. A third harmonic would mean that the frequency is 3 omega, thus the circuit will consist only of the quantities that have 3 omega in the expression. For example if you have : e(t)=5sin(3ωt+π/4) and is(t)=7sin(ωt+π/4), in the first harmonic the e(t) quantity will disappear from the circuit while in the third harmonic the is(t) quantity will be neglected.
What are two rules for the voltage and current in a parallel circuit?
f|_|ck that sh!t nig im looking for tat sh!t 2
If an RC Circuit is supplied with 24VDC and the circuit is in its third time constant how much voltage would be present across the capacitor?
If this is a homework assignment, please consider trying to answer it yourself first, otherwise the value of the reinforcement of the lesson offered by the assignment will be lost on you.Time constant is an exponential function, in some kind of form of e(-T/t) where T is time and t is time constant. If you evaluate this for T=3 and t=1, you get about 0.05. This means that the voltage across the capacitor will reach 95% of its final value after three time constants.Whether this is 95% of 24V or 5% of 24V depends on how the circuit is designed, but the most probable answer with the limited information given is about 23V.
What does R-T mean in Charger R-T?
R/T refers to Road and Track
What does t-r mean in a t-r stretch fabric?
Tear-resistant?
What does R T mean?
B r t in text abbreviation means be right there
What does b r t mean?
B r t in text abbreviation means be right there
Tr equals r what does r mean if t is not equal to 1?
r=0,Tr-r = 0 = r(T-1), since T != 1, then T-1 is non zero so r must be zero.
In the equine business what does T P R mean?
T= Temperature P= Pulse R= Respiration
What does o e r t o mean?
Painless operation.O E R T OoPerAtIoN
What does s m a r t mean?
Sexually Malested And Raped Twice S M A R T
What does c r t mean?
choose the right
What does D T R mean?
It means 'define the relationship'
What is mean by T R P?
Television rating point...
Why is energy lost when a capacitor is charged using a cell assuming resistance of connecting wires is zero?
Original answer: A capacitor will charge up from a DC at a rate dictated by the circuit RC constant. If R = 0, the capacitor would instantly charge to 100%, which obviously does not happen. the resistance in this circuit is the internal resistance of the battery, and the internal resistance of the capacitor (which is usually very low relative to the battery resistance). With this in mind, it should be clear that energy is "lost" by turning to heat due to the internal resistance of the battery (and the resistance of the capacitor). Expanded answer: The poster requested further explanation on why energy is "lost" due to turning to heat due to the resistance in the circuit. I'm using some very basic calculus below (simple integration). Let's start with some basic equations: Vc = the voltage across the capacitor at any given time = Vs (1 -- e^[-t/T]) Vs = the source voltage, which I'm assuming is a constant DC source I = the current in the series circuit at any given time= (Vs / R) e^[-t/T] Where t = time, and T is the circuit time constant dictated by R*C (resistance times capacitance of the circuit) Now I want to determine three things: 1). Total power over time (ie energy) from the source 2). Total power over time stored by the capacitor 3). Total power over time dissipated by the circuit resistance 1:Instantaneous power = V*I; total energy provided by the source is the integral of V*I: Esource = Integral (Vs) * (Vs/R)e^[-t/T] from time 0 to infinite Esource = (Vs^2)/R * (-T * e^[-t/T]) from t = 0 to t = infinite Esource = (Vs^2)/R * (-T) * (0 -- 1) = (Vs^2 / R) * T = C*Vs^2 2. Ecap = Vc*I = Vs(1 -- e^[-t/T]) * (Vs/R) e^[-t/T] from time 0 to infinite Ecap = (Vs^2/R) * (e^[-t/T] -- e^[-2t/T]) from time 0 to infinite (integrate each 'e' separately...) Ecap = (Vs^2/R) * (T -- T/2) = ½ C*Vs^2 3. The voltage across the resistance of the circuit will be whatever is "left over" that is not across the capacitance Eres = V*I = Vs e^[-t/T] * (Vs/R) e^[-t/T] from time 0 to infinite Eres = (Vs^2/R) e^[-2t/T] from time 0 to infinite Eres = (Vs^2/R) * (T/2) = ½ C*Vs^2 So ½ the energy used to charge a capacitor is "used up" by the Thevenin resistance (R) of the circuit. When power flows through a resistor, it is dissipated as heat. It does not matter if this resistance is very small, or very large, or mostly internal to the capacitor, or mostly internal to the battery source, the above 1-3 hold true for all cases.
