Two cells of unequal emfs, E1 and E2, the internal resistance r1 and r2 are joined as shown. VA and VB are the potentials at A and B respectively
A
one cell with continuously supply energy to the other
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B
the potential difference across both the cells will be equal
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C
the potential difference across one cell will be grater than its emf
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D
VA−VB=(E1r2−E2r1)(r1+r2)
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Solution
The correct options are A one cell with continuously supply energy to the other B the potential difference across both the cells will be equal C the potential difference across one cell will be grater than its emf Let i be the current flow in clockwise direction in the circuit . thus, E2−E1=i(r1+r2) or i=E2−E1r1+r2 now, VA−VB=E2+ir2=E2+(E2−E1)r2r1+r2=E2r1+E2r2+E2r2−E1r2r1+r2=E2r1+2E2r2−E1r2r1+r2 This is also potential across each cell. Thus, VA−VB>E2 Current flows in the cell of emf E2from the positive plate to the negative plate inside the cell and hence it absorbs energy. ans :A,B,C