A circuit consists of a resistor of R=7 Ω, a capacitor of C=7 μF, and an ideal self-inductor of L=0.04 H. All three are in series with a power supply that generates an EMF of 5sin(ωt) Volt. The internal resistance of both the power supply and the inductor are negligibly small. The system is at resonance.
(a) What is the time averaged power (in Watts) generated by the power supply?
We decrease the frequency of the power supply to a value for which the reactance (1ωC−ωL) becomes equal to R; the maximum EMF remains 5 V.
(b) What now will be the time averaged power (in Watts) generated by the power supply?
formula for 1 bit v^2/(2*R) v:maximum value
second answer half the 1 one
THANX!!
Pls. tell what is v and maximum value.
vrms
To find the time-averaged power generated by the power supply, we can use the formula:
P = (1/2) * Vrms * Irms * cos(θ)
where P is the power, Vrms is the root-mean-square voltage, Irms is the root-mean-square current, and θ is the phase angle between the voltage and the current.
In the given circuit, the power supply generates an EMF of 5sin(ωt) Volts, which means the root-mean-square voltage (Vrms) will be 5/sqrt(2) V. Since the system is at resonance, the reactance (1/ωC - ωL) becomes equal to R, which means the current will be in phase with the voltage (θ = 0).
Now, let's calculate the current (Irms):
Using Ohm's Law, we can calculate the impedance (Z) of the circuit using the formula:
Z = sqrt(R² + (X - XL)²)
where X is the reactance for the capacitor (1/ωC) and XL is the reactance for the inductor (ωL).
At resonance, X - XL = R, so the impedance simplifies to:
Z = sqrt(R² + R²) = sqrt(2R²) = R * sqrt(2)
The current (Irms) can be calculated by dividing the voltage (Vrms) by the impedance (Z):
Irms = Vrms / Z
Since Vrms = 5/sqrt(2) Volts and Z = R * sqrt(2) Ohms, we get:
Irms = (5/sqrt(2)) / (R * sqrt(2)) = 5 / (R * 2) = 5 / (7 * 2) = 5 / 14 Amperes
Finally, substituting the values into the power formula, we have:
P = (1/2) * (5/sqrt(2)) * (5/14) * cos(0)
= (1/2) * (5/sqrt(2)) * (5/14) * 1
= (25/2) * (1/sqrt(2)) * (1/14)
= (25/2) * (1/2sqrt(2)) * (1/7)
= 25 / (2 * 2 * sqrt(2) * 7)
= 25 / (4 * sqrt(2) * 7)
= 25 / (28 * sqrt(2))
≈ 0.375 Watts
Therefore, the time-averaged power generated by the power supply is approximately 0.375 Watts.