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Question

Power absorbed by an element for t = 10 sec, if the current magnitude is 2e-0.1t and the voltage across the element is \(V=6\frac{di}{dt}\), the absorbed power is :

A. - 0.325 Watts

B. - 2.4 e-0.2(10) Watts

C. 0.325 Watts

D. 2.4 e-0.2(10) Watts

E. 0.625 Watts

choose the correct answer from the options given below :

This question was previously asked in
UGC NET 2023 Electronic Science Question Paper (13-Dec-2023) (Shift 1)
The correct answer is

A & B only

To solve this problem, we need to determine the power absorbed by an element at \( t = 10 \) seconds, given the current and voltage expressions for the element. Let's go step-by-step:

Given the current \( i(t) = 2e^{-0.1t} \).

The voltage across the element is given by the expression: \(V = 6\frac{di}{dt}\).

Firstly, we need to find \(\frac{di}{dt}\):

\[\frac{di}{dt} = \frac{d}{dt}(2e^{-0.1t}) = 2 \times (-0.1) \times e^{-0.1t} = -0.2e^{-0.1t}\]

Substitute \(\frac{di}{dt}\) into the voltage equation:

\[V = 6 \times (-0.2e^{-0.1t}) = -1.2e^{-0.1t}\]

Now, calculate the power using the formula: \(P = V \cdot i\)

\[P = (-1.2e^{-0.1t})(2e^{-0.1t})\]

Simplifying this, we get:

\[P = -2.4e^{-0.2t}\]

Substitute \( t = 10 \) into the power equation to find the power at that specific time:

\[P = -2.4e^{-0.2 \times 10} = -2.4e^{-2} \]\]

Calculate the above expression to find the exact value:

\(-2.4e^{-2} = -2.4 \times 0.1353 \approx -0.325 \text{ Watts}\)

Therefore, the correct options corresponding to these calculations are:

  • -0.325 Watts
  • -2.4e^{-0.2(10)} Watts

Conclusion: The correct answer choices are A & B only as they represent both forms of the expression calculated for the power absorbed.

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