Question

✇ The series and parallel circuits shown in figure have the same impedance and the same power factor. If R=3 Ω and X = 4 Ω, find the values of $\sf R_1$, and $\sf X_1$. Also, find the impedance and power factor.


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Answer 1

Answer :

Both series and parallel circuits have same impedance and the same power factors.

In series connection,

• R = 3Ω
• X = 4Ω

We have to find values of R₁ and X₁$.$

★ Question is not really tough! Just solve step by step. In order to solve this question you should be familiar with following things!

• How to calculate impedance of RC, LC, RL or LCR circuit.
• How to calculate power factor of AC circuit. (also convert cos∅ into tan∅ in such questions)

That's all! Now let's solve this question.

⧪ Impedance of circuit :

⭆ Z = √(R² + X²)

⭆ Z = √(3² + 4²)

⭆ Z = √25

⭆ Z = 5Ω

⧪ Power factor of circuit :

⭆ Power factor (cos∅) = R/Z

⭆ cos∅ = 3/5

So value tan∅ will be 4/3.

Now let's solve second part of question!

• For resistor containing circuit, phase difference between voltage and current = zero
• For inductor containing circuit, phase difference between voltage and current = 90°

Refer to the attachment for phase diagram.

Let current flow in resistor be I₁ and that in inductor be I₂.

Remember that we can't add I₁ and I₂ directly as there is difference in phase so we have to use vector method in order to add I₁ and I₂.

➠ I² = I₁² + I₂²

• current = voltage/resistance

Voltage remains same in parallel so it will be cancel from from sides.

➠ 1/Z² = 1/R₁² + 1/X₁² ... (I)

From the attached diagram,

• tan∅ = R₁/X₁ = I₂/I₁ = 4/3 ... (II)

Therefore, R₁ = 4X₁/3 and Z = 5

Substituting value of R₁ in (I), we get

➠ 1/25 = 1/X₁² + 9/16X₁²

➠ 1/25 = 1/X₁² (25/16)

➠ 1/X₁² = 16/(25)²

➠ 1/X₁ = 4/25

➠ X₁ = 25/4Ω

Substituting value of X₁ in (II), we get

➠ X₁/R₁ = 3/4

➠ R₁ = 4X₁/3

➠ R₁ = 4/3(25/4)

➠ R₁ = 25/3Ω

Answer 2

Both series and parallel circuits have same impedance and the same power factors.

In series connection,

R = 3Ω

X = 4Ω

We have to find values of R₁ and X₁

⧪ Impedance of circuit :

⭆ Z = √(R² + X²)

⭆ Z = √(3² + 4²)

⭆ Z = √25

⭆ Z = 5Ω

⧪ Power factor of circuit :

⭆ Power factor (cos∅) = R/Z

⭆ cos∅ = 3/5

So value tan∅ will be 4/3.

Now let's solve second part of question!

For resistor containing circuit, phase difference between voltage and current = zero

For inductor containing circuit, phase difference between voltage and current = 90°

Refer to the attachment for phase diagram.

Let current flow in resistor be I₁ and that in inductor be I₂.

Remember that we can't add I₁ and I₂ directly as there is difference in phase so we have to use vector method in order to add I₁ and I₂.

➠ I² = I₁² + I₂²

current = voltage/resistance

Voltage remains same in parallel so it will be cancel from from sides.

➠ 1/Z² = 1/R₁² + 1/X₁² ... (I)

From the attached diagram,

tan∅ = R₁/X₁ = I₂/I₁ = 4/3 ... (II)

Therefore, R₁ = 4X₁/3 and Z = 5

Substituting value of R₁ in (I), we get

➠ 1/25 = 1/X₁² + 9/16X₁²

➠ 1/25 = 1/X₁² (25/16)

➠ 1/X₁² = 16/(25)²

➠ 1/X₁ = 4/25

➠ X₁ = 25/4Ω

Substituting value of X₁ in (II), we get

➠ X₁/R₁ = 3/4

➠ R₁ = 4X₁/3

➠ R₁ = 4/3(25/4)

➠ R₁ = 25/3Ω