Kirchhoff's law

Question

Given: !!V_a = 2V, V_b= 3V,V_c= 6V!!, !!I-R!! graph will be:

(a) !!I_(max)= 6/R_0!!

(b) !!I_(max)= 5/R_0!!

(c) !!I_(max)= 10/R_0!!

(d) !!I_(max)= 8/R_0!!

Solution

Suppose current in the circuit is, !!I!!

Apply Kirchoff's voltage law,

!!IR+IR_0- V_a+V_b-V_c = 0!!

!!I(R+R_0)= 5!!

!!=> I = 5/(R+R_0)!!

!!I_(max)!!, when !!R=0!!

Therefore,

!!I_(max) = 5/R_0!!

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