MHT CET · Physics · Alternating Current

When 80 volt d.c. is applied across a solenoid, a current of 0.8 A flows in it. When 80 volt a.c. is applied across the same solenoid, the current becomes 0.4 A . If the frequency of a.c. source is 50 Hz , the impedance and inductance of the solenoid is nearly

A \(200 \Omega, 0.55 \mathrm{H}\)
B \(100 \Omega, 0.8 \mathrm{H}\)
C \(300 \Omega, 1.2 \mathrm{H}\)
D \(200 \Omega, 1.5 \mathrm{H}\)

Verified Solution

Answer & Solution

Correct Answer

(A) \(200 \Omega, 0.55 \mathrm{H}\)

Step-by-step Solution

Detailed explanation

When 80 Vdc is applied,
\(Z =\sqrt{\mathrm{R}^2+(\omega \mathrm{L})^2}=\mathrm{R} \quad \ldots(\because \omega=\) \(0 \text { in dc circuit })\)
\(\therefore \mathrm{Z} =\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{80}{0.8}=100 \Omega \quad \ldots \text { (i) }\)
When 80 V ac is applied,
\(\begin{aligned}
& Z=\frac{V}{I}=\frac{100}{0.5}=200 \Omega ...(ii)\\
& Z=\sqrt{R^2+(\omega L)^2} \\
& \Rightarrow(2 \pi \mathrm{fL})^2=Z^2-R^2 \\
& L=\sqrt{\frac{200^2-100^2}{(2 \pi \times 50)^2}}=0.55 \mathrm{H}
\end{aligned}\)
...[From(i).and (ii)]

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.

Same subject

Explore more questions on app

From MHT CET

Explore more questions on app