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Sloan [31]
4 years ago
15

I need help doing what the difference between the symbol and the number​

Physics
1 answer:
fomenos4 years ago
3 0

Answer:

Number :

              Number is actually represents quantity.

Symbol :

              Symbol is actually a character that shows the number.

Explanation:

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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:

brainly.com/question/4438943

brainly.com/question/10597501

brainly.com/question/12246020

#LearnwithBrainly

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