Question

A source of voltage 50V, 25 Hz is connected
in series to a resistance of 4000, and
inductance of H, and a capacitor of
10 uF. What is the average power
dissipated in the circuit (in W) ?​

Answer 1

Answer:

Average Power dissipated = 0.307978 Watt

Explanation:

Given that:

Source Voltage =50 V

rms voltage = 50/√2 = 35.35 volt

Frequency of AC source(f) = 25 Hz

So, Angular frequency = ω = 2πf = 50π

Resistance in circuit (R) = 4000 Ohm

Inductance (L) = H henry

Capacitance (C)= 10 micro farad = 10 × 10^-6 F

So, Reactance of Capacitor =

$Xc = \frac{1}{ωc} = \frac{1}{50 π \times {10}^{ - 5} } = 636.62 ohm$

Similarly Reactance for capacitor =

$Xl = ω \times L= 50 π \times \: H =157 H \: ohm$

Total Impedence of the AC circuit (Due to R, C, and L):

$z = \sqrt{ {r}^{2} + {( xc - xl)}^{2} }$

$z = \sqrt{ {4000}^{2} + {( 636.65- 157)}^{2} } ohm \\ \\ z = 4028.65 \: ohm$

Average Power dissipated in Circuit;

$w = {i \: rms}^{2} \times r \\ = { (\frac{v \: rms}{z} )}^{2} \times r$

$w = { (\frac{35.35}{4028.65} )}^{2} \times 4000 \: watt \\ w = 0.307978 \: watt$

Answer 2

Answer:

The average power dissipated in the circuit (in W) is 1.25.

Explanation:

Given,

The voltage of the source is (V) = 50 V

The frequency of the battery (f) = 25 Hz.

The resistance of the resistance (R) = 4000 ohms.

The inductance of the inductor is 1H.

The capacitance of a capacitor (C) = 10 μF = 10⁻⁵ F

To find,

The average power dissipated in the circuit.

Calculation,

R.M.S voltage ($V_{rms}$) = V/√2 = 50/√2 = 35.35 V.

Now, Frequency of AC source(f) = 25 Hz

So, Angular frequency = ω = 2πf = 50π

The reactance of a capacitance C is given by:

$X_c = \frac{1}{\omega C} = \frac{1}{50\pi \times 10^{-5}}$ = 636.62 ohms.

Similarly, the reactance of an inductor is given by:

$X_L = \omega L$ = 50π × 1H = 157 ohm.

Now the total impedance(Z) can be calculated as:

$Z = \sqrt{R^2 + (X_C - X_L)^2}\\ \implies Z = \sqrt{4000^2 + (636.62 - 157)^2}$

⇒ Z = 4028.65 ohm.

Average Power dissipated in Circuit:

$P = (i_{rms})^2 \times R\\\implies P = (\frac{V_{rms}}{z})^2 \times R \\\implies P = (\frac{35.35}{4028.65} )^2 \times 4000$

⇒ P = 0.307 Watts

Therefore, the average power dissipated in the circuit (in W) is 0.307.

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