\(LR\) શ્રેણી પરિપથને \(\omega \) આવૃતિ ધરાવતા \(AC\) સ્ત્રોત સાથે જોડેલ છે જેનો ઇન્ડક્ટિવ રીએક્ટન્સ \(2R\) છે. હવે \(R\) કેપેસિટીવ રીએક્ટન્સ ધરાવતા કેપેસીટરને \(L\) અને \(R\) સાથે શ્રેણીમાં જોડેલ છે. તો નવા અને જૂના પાવર ફેકટરનો ગુણોત્તર શું મળે?

A \(\sqrt {\frac{2}{3}} \)
B \(\sqrt {\frac{2}{5}} \)
C \(\sqrt {\frac{3}{2}} \)
D \(\sqrt {\frac{5}{2}} \)

Verified Solution

Answer & Solution

Correct Answer

(D) \(\sqrt {\frac{5}{2}} \)

Step-by-step Solution

Detailed explanation

Power factor \(_{(old)}\) \(=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^{2}+\mathrm{X}_{\mathrm{L}}^{2}}}=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^{2}+(2 \mathrm{R})^{2}}}=\frac{\mathrm{R}}{\sqrt{5 \mathrm{R}}}\) Power factor \(_{(\text {new })}\)…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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