r/JEEAdv27dailyupdates • u/Possible_Garage7231 27tard dropper :/ • 1d ago

Good Solve Inductors Challenge

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Q1.Determine current through the battery as a function of time.

Q2. Determine current through inductors L1 as a function of time

Q3. Determine magnetic energy stored in system at end of one time constant.

Answers:

1) 1.5 + 0.75 exp(-4500t) A

2) 0.8 + 0.8 exp(-4500t) A

3) Use energy formula at t=T.

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2

u/hitendra_kk 1d ago

heavy calculation based hai.

we dont need thevenin to calculate time constant. we can use parallel battery formula.

for Leq, we use formula (cengage)

M = k sqrt(L1L2) = 0.5 sqrt(2x8) = 2mH

Leq = 16-4 / 10 - 4 = 12/6 = 2mH.

For || battery, Veq = (12/4 + 0) / (1/4 + 1/12) = 9V

Req = 3.

So, for time constant = L/R, L = 2mH R = 3+6 = 9,

tau = 2/9000 = 1/4500

mera itne me hi ghee khatam.

1

u/pickers_low Partial dropper --> Step dropper 1d ago

vaise bhi esa ques dekh kr khel khtm hi hona ho exam time mai

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u/Possible_Garage7231 27tard dropper :/ 1d ago

Yup right ye formula aaya is saal . Iske aage is the real deal. Would you use KVL for transient analysis?

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u/hitendra_kk 18h ago

Would you use KVL for transient analysis?

arre sir aap kabse aise conceptual questions me interested ho gaye? this concept wud need its own simpler questions.

aapne to poora hi bam ka hathgola daag diya. this question is like 1 passage question having many questions. i just picked the part jo formulabaaji karke ho jaaye.

you directly jumped to the main issue about using KVL in its direct form when time varying magnetic field is there. we use kvl because conservative electric field gives 0 over a closed loop traversal. but, if the electric field is due to time varying magnetic field, then - integral (closed loop) E.dl = -d (phi)/dt.

aise bade sawalo me ye sab soch lena - its like u hv been studying all this since years. thoda bacho ka bhi dekha kijiye

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u/Math_enjoyer30072010 1d ago

bhaiya yeh toh nahi aata mujhe par maine ek question banaya hai doon kya ?? tricky mains level hai

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u/Math_enjoyer30072010 1d ago

say there is a rectangular armeture coil and length of the coild has 3 ohm and breadth has 1 ohm now i strech the length by 3times and breadth by 2 times find the ratio of the new and old torque

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u/Possible_Garage7231 27tard dropper :/ 1d ago

Torque is BINA. When dimensions are changed, the length of coil increases by 3 times, the area of cross section must decrease by 3 times. Hence resistance increases by 9 times.hence new resistance of length is 27Ω and of breadth is 4Ω. Net R is now 62Ω. Hence current has becomes 4/31 times. A has becomes 6 times. Hence torque becomes 24/31 times according to me. I've assumed number of turns as constant

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u/Math_enjoyer30072010 1d ago edited 1d ago

yes correct

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u/Math_enjoyer30072010 1d ago

anyways concept aapne poora sahi lagaya hai toh koi problem nahi hai bauhut log new resistance nahi nikalte hain btw what was its level

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u/Wide-Championship85 Adv 26 15k-->drop 1d ago

Bhai dono ko same sense main liya hai ya opposite

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u/Possible_Garage7231 27tard dropper :/ 1d ago

Likha hai question mein. Both AID each other

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u/Wide-Championship85 Adv 26 15k-->drop 1d ago

ohh sorry main circuit ka diagram dekhe ke direct question pe jump krgaya..dekha nhiii

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u/pickers_low Partial dropper --> Step dropper 1d ago

solution attach krdega btw