Streamlined objects are smooth and ?

A 67 Vrms source is placed across a load that consists of a 12 ohm resistor in series with an capacitor whose reactance is 5 ohm

SVETLANKA909090 [29]

a) The average true power is 318.3 W

b) The reactive power is 132.6 W

c) The apparent power is 344.8 W

d) The power factor is 0.92

Explanation:

a)

For a circuit made of a resistor and a capacitor, the average (true) power is given by the resistive part of the circuit only.

Therefore, the average true power is given by:

$P=I^2R$

where

I is the current

R is the resistance

In this problem, we have

V = 67 V (rms voltage)

$R=12 \Omega$ is the resistance of the load

$X=5\Omega$ is the reactance of the circuit

First we have to find the impedance of the circuit:

$Z=\sqrt{R^2+X^2}=\sqrt{12^2+5^2}=13 \Omega$

Then we can find the current in the circuit by using Ohm's law:

$I=\frac{V}{Z}=\frac{67}{13}=5.15 A$

Therefore, the average true power is

$P=I^2R=(5.15)^2(12)=318.3 W$

b)

The reactive power of a circuit consisting of a resistor and a capacitor is the power given by the capacitive part of the circuit.

Therefore, it is given by

$Q=I^2X$

where

I is the current

X is the reactance of the circuit

In this circuit, we have

$I=5.15 A$ (current)

$X=5 \Omega$ (reactance)

Therefore, the reactive power is

$Q=(5.15)^2(5)=132.6W$

c)

In a circuit with a resistor and a capacitor, the apparent power is given by both the resistive and capacitive part of the circuit.

Therefore, it is given by

$S=I^2Z$

where

I is the current

Z is the impedance of the circuit

Here we have

I = 5.15 A (current)

$Z=13 \Omega$ (impedance)

Therefore, the apparent power is

$S=I^2 Z=(5.15)^2(13)=344.8 W$

d)

For a circuit with a resistor and a capacitor, the power factor is the ratio between the true power and the apparent power. Mathematically:

$PF=\frac{P}{S}$

where

P is the true power

S is the apparent power

For this circuit, we have

P = 318.3 W (true power)

S = 344.8 W (apparent power)

So, the power factor is

$PF=\frac{318.3}{344.8}=0.92$

Learn more about power and circuits:

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Consider two soap bubbles with radius r1 and r2 (r1 <r2) connected via a valve. What happens if we open the valve

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Complete Question

The complete question is shown on the first uploaded image

Answer:

The pressure difference of the first bubble is $\Delta P _1 =10 J/m^3$

The pressure difference of the second bubble is $\Delta P _2 =20 J/m^3$

The pressure difference on the second bubble is higher than that of the first bubble so when the valve is opened pressure from second bubble will cause air to flow toward the first bubble making is bigger

Explanation:

From the question we are told that

The radius of the first bubble is $r_1 = 10 \ mm=0.01 \ m$

The radius of the second bubble is $r_2 = 5 \ mm = 0.005 \ m$

The surface tension of the soap solution is $s = 25 \ mJ/m^2 = 25*10^{-3} J/m^2$

Generally according to the Laplace's Law for a spherical membrane the pressure difference is mathematically represented as

$\Delta P = \frac{4 s}{R}$

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$\Delta P _1 = \frac{4 s}{r_1}$

substituting values

$\Delta P _1 = \frac{4 *25 *10^{-3}}{0.01}$

$\Delta P _1 =10 J/m^3$

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$\Delta P _2 = \frac{4 s}{r_1}$

$\Delta P _2 = \frac{4 *25 *10^{-3}}{0.005}$

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