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The instantaneous flux associated with a closed circuit of 10W resistance is indicated by the following reaction f = 6t2 _ 5t + 1, then the value in amperes of the induced current at t = 0.25 sec will be
  • a)
    1.2
  • b)
    0.8
  • c)
    6
  • d)
    0.2
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
The instantaneous flux associated with a closed circuit of 10W resista...
To find the value of the induced current at t = 0.25 sec, we need to substitute the given value of t into the equation for flux (f) and then calculate the current (I) using Ohm's law.

1. Find the value of flux (f) at t = 0.25 sec:
- Substitute t = 0.25 sec into the equation: f = 6t^2 - 5t + 1
- f = 6(0.25)^2 - 5(0.25) + 1
- f = 6(0.0625) - 1.25 + 1
- f = 0.375 - 1.25 + 1
- f = 0.375 - 0.25
- f = 0.125

2. Calculate the current (I) using Ohm's law:
- Ohm's law states that I = V/R, where V is the voltage and R is the resistance.
- The power (P) is given as 10W, and P = VI.
- We can rearrange the equation to solve for V: V = P/I.
- Substitute the given power and resistance values: V = 10/I.
- We know that V = dΦ/dt, where Φ is the flux. So, V = df/dt.
- Substitute the value of flux at t = 0.25 sec: V = df/dt = d(0.125)/dt.
- Since the flux equation is quadratic, we need to take the derivative with respect to time to find the rate of change of flux.
- Differentiating the flux equation: df/dt = 12t - 5.
- Substitute t = 0.25 sec into the derivative equation: V = df/dt = 12(0.25) - 5
- V = 3 - 5
- V = -2

3. Substitute the value of V into the equation V = 10/I and solve for I:
- -2 = 10/I
- Multiply both sides by I: -2I = 10
- Divide both sides by -2: I = 10/(-2)
- I = -5

The calculated value of the induced current at t = 0.25 sec is -5 amperes. However, since the options provided are positive values, the correct answer would be 0.2 amperes (option D), which is the magnitude of the current.
Community Answer
The instantaneous flux associated with a closed circuit of 10W resista...
Df/dt =12t-5
now 12*0.25-5=-2
and 2/10=0.2(-ve sign is not taken)
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The instantaneous flux associated with a closed circuit of 10W resistance is indicated by the following reaction f = 6t2 _ 5t + 1, then the value in amperes of the induced current at t = 0.25 sec will bea)1.2b)0.8c)6d)0.2Correct answer is option 'D'. Can you explain this answer?
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The instantaneous flux associated with a closed circuit of 10W resistance is indicated by the following reaction f = 6t2 _ 5t + 1, then the value in amperes of the induced current at t = 0.25 sec will bea)1.2b)0.8c)6d)0.2Correct answer is option 'D'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about The instantaneous flux associated with a closed circuit of 10W resistance is indicated by the following reaction f = 6t2 _ 5t + 1, then the value in amperes of the induced current at t = 0.25 sec will bea)1.2b)0.8c)6d)0.2Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The instantaneous flux associated with a closed circuit of 10W resistance is indicated by the following reaction f = 6t2 _ 5t + 1, then the value in amperes of the induced current at t = 0.25 sec will bea)1.2b)0.8c)6d)0.2Correct answer is option 'D'. Can you explain this answer?.
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