InterviewSolution
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InterviewSolution
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Two identical cells , whether joined together in series or in parallel give the same current, when connected to external resistance of `1 Omega`. Find the internal resistance of each cell. |
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Answer» Let `epsilon`, r be the emf and internal resistance of each cell. External `R= 1 Omega`. When the two cells are connected in series Total emf of cells `= epsilon + epsilon = 2 epsilon` Total resistance of circuit `= R +r+r=1+2r` Currnet in the circuit, `I_(1)= (2epsilon)/(1+2r)` When the two cells are connseted in parallel Effective emf of two cells = emf of single cell=`epsilon` Total internal resistance of two cells `(r xx r)/(r+r)=r/2` Total resistance of the circuit `= R + r//2 = 1+r//2`. Current in the circuit, `I_(2)=(epsilon)/(1+r//2)` As per question, `I_(1) = I_(2)` So, `(2 epsilon)/(1+2r) = (epsilon)/(1+r//2) = (2 epsilon)/(2+r)` or `1 + 2r = 2+r or r = 1 Omega` |
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