%3Cp%3E%3Cspan%20class%3D%22c1%22%3EGiven%20the%20linear%20circuit%20in%20the%20

Question

%3Cp%3E%3Cspan%20class%3D%22c1%22%3EGiven%20the%20linear%20circuit%20in%20the%20figure%20below%2C%0Ait%20is%20known%20that%20when%20a%204%20k%3F%20load%20is%20connected%20to%20the%20terminals%0AA-B%2C%20the%20load%20current%20is%200.522%20mA.%20If%20a%2015%20k%3F%20load%20is%20connected%20to%0Athe%20terminals%2C%20the%20load%20current%20is%200.353%20mA.%20Find%20the%20current%20in%20a%0A26%20k%3F%20load.%3C%2Fspan%3E%3C%2Fp%3E%0A%3Cp%3E%3Cimg%20class%3D%22user-upload%22%20src%3D%0A%22http%3A%2F%2Fmedia.cheggcdn.com%2Fmedia%252F0a8%252F0a8089a7-4ece-4a17-ad78-ebfcc576b377%252FphpodkU4N.png%22%0Aheight%3D%22350%22%20width%3D%221257%22%20%2F%3E%3C%2Fp%3E%0A

Explanation / Answer

Vth = I(Rth+Rl)

Vth = 0.522(Rth+4)

Vth=0.353(Rth+15)

0.169Rth = 0.353*15 - 0.522*4

Rth = 18.97Kohm

Vth = 0.522(18.97+4) = 12V

I = Vth/(Rth+26) = 12/(18.97+26) = 0.2668mA

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