A) 1.5 A
B) 3.0 A
C) 15 A
D) 30 A
Correct Answer: A
Solution :
Given problem is the case of mixed grouping of cells So total current produced \[i=\frac{nE}{R+\frac{nr}{m}}\] Here \[m=100,\,\,n=5000,\,\,R=500\,\Omega \] \[E=0.15\,V\] and \[r=0.25\,\Omega \] \[i=\frac{5000\times 0.15}{500+\frac{5000\times 0.25}{100}}\]\[=\frac{750}{512.5}\] \[\approx 1.5\,A\]You need to login to perform this action.
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