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JEE Main 20206 SeptemberMorning Shift
Exam
JEE Main
Year
2020
Date
6 September
Shift
Morning Shift
Subject
Physics
Chapter
Alternating Current

JEE Main 2020 · 6 September · Morning Shift · Physics · Alternating Current

A part of a complete circuit is shown in the figure. At some...

A part of a complete circuit is shown in the figure. At some instant, the value of current I is 1A1A and it is decreasing at a rate of 102As−1{10}^{2}{As}^{- 1}. The value of the potential difference Vp−VQ,{V}_{p} - {V}_{Q}, (in volts) at that instant is-

A part of a complete circuit is shown in the figure. At some instant, the value of current I is 1A and it is decreasing

Correct answer

0

Step-by-step solution

Vp−Ldldt−30+RI=VQ{V}_{p} - L\dfrac{dl}{dt} - 30 + RI = {V}_{Q}

VP−Va=+50×10−3×102+30−1×2{V}_{P} - {V}_{a} = + 50 \times {10}^{- 3} \times {10}^{2} + 30 - 1 \times 2

=+5+30–2=33= + 5 + 30 – 2 = 33

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