ધારો કે બે નજીકની કુંડળીઓ 1 અને 2 માંથી અનુક્રમે \(\mathrm{I}_1\) અને \(\mathrm{I}_2\) પ્રવાહો એકસાથે વહે છે. જો કુંડળી \(1\) નું આત્મપ્રેરકત્વ \(\mathrm{L}_1=\) અને કુંડળી 2 ના સંદર્ભમાં કુંડળી 1 નું અન્યોન્ય પ્રેરકત્વ \(\mathrm{M}_{12}=\), તો કુંડળી 1 માં પ્રેરિત EMF નું મૂલ્ય ________ હશે.

A \(\varepsilon_1=-L_1 \frac{\mathrm{dI}_2}{d t}-\mathrm{M}_{12} \frac{\mathrm{dI}_1}{d t}\)
B \(\varepsilon_1=-L_1 \frac{\mathrm{dI}_1}{d t}-\mathrm{M}_{12} \frac{\mathrm{dI}_2}{d t}\)
C \(\varepsilon_1=-L_1 \frac{\mathrm{dI}_1}{d t}-\mathrm{M}_{12} \frac{\mathrm{dI}_1}{d t}\)
D \(\varepsilon_1=-L_1 \frac{\mathrm{dI}_1}{d t}+\mathrm{M}_{12} \frac{\mathrm{dI}_2}{\mathrm{dt}}\)

Verified Solution

Answer & Solution

Correct Answer

(B) \(\varepsilon_1=-L_1 \frac{\mathrm{dI}_1}{d t}-\mathrm{M}_{12} \frac{\mathrm{dI}_2}{d t}\)

Step-by-step Solution

Detailed explanation

\(\begin{aligned} & \phi_1=\mathrm{L}_1 \mathrm{I}_1+\mathrm{M}_{12} \mathrm{I}_2 \\ & \varepsilon_1=-\frac{\mathrm{d} \phi_1}{\mathrm{dt}}=-\mathrm{L}_1 \frac{\mathrm{dI}}{1} \frac{\mathrm{I}_{12}}{\mathrm{dt}}-\mathrm{M}_{12} \frac{\mathrm{~d} I_2}{\mathrm{dt}}\end{aligned}\)

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 JEE Mains

Explore more questions on app