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This is solution to problems related Electrical Engineering course. It was suggested by Prof. Bhooshan Sawhney at Shree Ram Swarup College of Engineering and Management. It includes: Circular, Plate, Central, Discharge, Current, Magnetic, Field, Maximum, Result, Electric, Lines
Typology: Exercises
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A a =^
πR^2 π(R/2)^2 = 4^.
Thus, from Eqs. 32-38and 32-39 the total discharge current is given by i = id = 4(2.0 A) = 8.0 A.