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The potential difference across a 150mH inductor as a function of time is shown in figure. Assume that the initial value of the current in the inductor is zero. What is the current whent=2.0msandt=4.0ms?
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Answer
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Hint: The voltage across the inductor is directly proportional to the rate of change of current through the inductor. When applied voltage is variable then integration logic is applied for solving such types of questions and we get current is directly proportional to the area of the Voltage-time graph.

Complete step-by-step solution:
The V - I equation of the inductor is given as:
-VL=Ldidtequation 1
Equation 1 can be written as,
di=1LVLdt- - - - - - - - - - -Equation 2
As it is given in the question at time t = 0, current i = 0 and so let us assume at after time t current becomes i.
So on integrating above equation 2 from time 0 to t and current 0 to i , we get ,
0idi=1L0tVLdt
i=1L0tVLdt
According to the above equation current through the inductor is equal to the1L times the area under VL-time graph.
i=1L(Area_under_VLt_grpah) - - - - - - Equation 3
Now we calculate current for t=2.0ms, then half the graph is drawn.

  
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Now we have to find the current through the inductor at t=2.0ms then we have to substitute the values in equation 3, we get
i=1150×103(12×2×103×5)
On Solving we get ,
i=3.33×103A
So, the value of current at t=2.0ms is 3.33×103A
Now fort=4.0ms, full graph is drawn
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Similarly, to find the current through the inductor at t = 4.0ms we have to substitute the values below in equation 3.
i=1150×103(12×4×103×5)i=6.67×103A
So, the value of current at t=4.0ms is6.67×103A.
So, we can conclude that current at t=2.0ms and at t=4.0ms is 3.33×103Aand 6.67×103A respectively.

Note: Voltage in an inductor can only be calculated in the presence of changing current because changing current is having frequency. In the case of constant current passing through the inductor, the voltage is equal to zero and the inductor looks like a short circuit element.