sagda_m2
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This question is about the self induction:
The resistance of a coil is 15 ohm, itself induction is 0.6 Henry, it is connected to a source of direct current which gives 120 volts. Calculate the rate of growth of the current at the moment when the current reaches 80 % of its maximum value.
My Answer:
First I calculated the maximum current:
I= V/R = 120/5 =8 Ampere
Therefore: the 80 % of the maximum current =6.4 A
Then the electromotive force will be:
V= I * R = 6.4 * 15 = 96 volt
And since: the induced e.m.f in the coil = -L * (dI/dt)
Therefore: 96 = 0.6 * (dI/dt)
Therefore: dI/dt = 96/0.6 = 160 A/s
But…..
I found the answer in my book as follows:
Since e.m.f = -L*(dI/dt)
Therefore: (120-96) = 0.6 * (dI/dt)
Therefore: dI/dt = 24/0.6 = 40 A/s
So, why did he substitute the e.m.f in the equation with the difference in volts (120-96)? Why didn't he substitute with the (96 volts)?
Thanks in advance,
The resistance of a coil is 15 ohm, itself induction is 0.6 Henry, it is connected to a source of direct current which gives 120 volts. Calculate the rate of growth of the current at the moment when the current reaches 80 % of its maximum value.
My Answer:
First I calculated the maximum current:
I= V/R = 120/5 =8 Ampere
Therefore: the 80 % of the maximum current =6.4 A
Then the electromotive force will be:
V= I * R = 6.4 * 15 = 96 volt
And since: the induced e.m.f in the coil = -L * (dI/dt)
Therefore: 96 = 0.6 * (dI/dt)
Therefore: dI/dt = 96/0.6 = 160 A/s
But…..
I found the answer in my book as follows:
Since e.m.f = -L*(dI/dt)
Therefore: (120-96) = 0.6 * (dI/dt)
Therefore: dI/dt = 24/0.6 = 40 A/s
So, why did he substitute the e.m.f in the equation with the difference in volts (120-96)? Why didn't he substitute with the (96 volts)?
Thanks in advance,