(ब) 800 फेरों वाली एक कुण्डली में 1.5 एम्पियर की विद्युत धारा प्रवाहित करने पर
उसके प्रति फेरे में चुम्बकीय फ्लक्स का मान 1.5x10-5 वेबर हैं। कुण्डली
का स्वप्रेरकत्व ज्ञात कीजिए।
[1+2-31
(b) A current of 1.5 amperes in a 800 turns coil produces
1.5 * 10-5 weber magnetized flux in each turn. Find out the
self-inductance of coil.
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Explanation:
Given A current of 1.5 amperes in a 800 turns coil produces
1.5 * 10-5 weber magnetized flux in each turn. Find out the
self inductance of coil.
- So we need to find the self inductance of the coil.
- Given: Current I = 1.5 Amp
- Number of turns N = 800
- Magnetized flux φ = 1.5 x 10^-5 wb
- So for a coil Inductance will be
- L = Nφ / I
- = 800 x 1.5 x 10^-5 / 1.5
- = 1200 x 10^-5 / 1.5
- = 800 x 10^-5 henry
- Or L = 8 m h
Reference link will be
https://brainly.in/question/41270442
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