© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the Known Quantities
• Source Voltage, V_{\text{rms}} = 250\text{ V}
• Frequency, f = 50\text{ Hz}
• Power Dissipated, P = 400\text{ W}
• Power Factor of LR Circuit, \cos \phi = 0.8
Step 2: Relate Power Factor to Circuit Quantities
In a series LR circuit, the power factor is given by
\cos \phi = \frac{R}{Z},
where R is the resistance and Z is the total impedance of the circuit.
Step 3: Use Power Formula to Find the Impedance Z
The power in a series circuit can also be expressed as:
P = \frac{V_{\text{rms}}^{2}}{Z^{2}} \times R.
Substituting P = 400\text{ W} and V_{\text{rms}} = 250\text{ V} , we get
400 = \frac{(250)^{2}}{Z^{2}} \times R.
Since \cos \phi = \frac{R}{Z} = 0.8 , we have R = 0.8Z . Substitute R = 0.8Z into the power equation:
400 = \frac{(250)^{2}}{Z^{2}} \times 0.8Z \quad \Rightarrow \quad 400 = \frac{62500 \times 0.8}{Z} \quad \Rightarrow \quad 400Z = 62500 \times 0.8.
Solve for Z :
Z = \frac{62500 \times 0.8}{400} = 125\,\Omega.
Step 4: Determine the Resistance R and Inductive Reactance X_{L}
Since R = 0.8Z ,
R = 0.8 \times 125 = 100\,\Omega.
In a series LR circuit, the impedance is given by
Z = \sqrt{R^{2} + X_{L}^{2}}.
Hence,
125 = \sqrt{(100)^{2} + X_{L}^{2}}.
Solve for X_{L} :
125^{2} = 100^{2} + X_{L}^{2} \quad \Rightarrow \quad 15625 = 10000 + X_{L}^{2}
\quad \Rightarrow \quad X_{L}^{2} = 5625 \quad \Rightarrow \quad X_{L} = 75\,\Omega.
Step 5: Condition for Power Factor Unity (Resonance)
To bring the power factor to unity by adding a capacitor in series, we need
X_{L} = X_{C},
where X_{C} is the capacitive reactance. For a capacitor,
X_{C} = \frac{1}{\omega C} = \frac{1}{2 \pi f \, C}.
Given that X_{L} = 75\,\Omega , set X_{C} = 75\,\Omega .
Step 6: Solve for the Required Capacitance C
75 = \frac{1}{2\pi \times 50 \times C}.
Rearrange to find C :
C = \frac{1}{75 \times 2\pi \times 50}.
Compute this value:
C = \frac{1}{75 \times 100\pi}
= \frac{1}{7500\pi}\,\text{F}.
Converting to microfarads ( \mu\text{F} ), we multiply by 10^{6} :
C = \frac{10^{6}}{7500\pi}\,\mu\text{F}
= \frac{1000}{7.5\pi}\,\mu\text{F}
= \frac{1000}{7.5} \times \frac{1}{\pi}\,\mu\text{F}.
Step 7: Express Capacitance in the Given Form
The problem states that
C = \left(\frac{n}{3\pi}\right)\,\mu\text{F}.
We have
\frac{1}{75 \times 2\pi \times 50} = \frac{n \times 10^{-6}}{3\pi}.
Numerically solving gives
n = 400.
Final Answer
The value of n is 400.
