A) 20.4eV
B) 24.4eV
C) 19.8eV
D) 16.1eV
Correct Answer: A
Solution :
Key Idea: When current is drawn potential difference is less than emf. The emf (E) is the characteristic of each cell and its value remains constant for the cell, while the potential difference (V) goes on, decreasing on taking more and more current (i) from the cell. So, \[V=E-ir\] where r is internal resistance of cell. Given, \[{{V}_{1}}=20\,V,\,{{i}_{2}}=2\,A,\,{{i}_{1}}=0.2\,A\] First case: \[20=E-0.2\,r\] \[\Rightarrow \] \[E=20=0.2\,r\] ... (i) Second case: \[16=E-2\,r\] \[\Rightarrow \] \[E=16+2\,r\] ... (ii) Solving Eqs. (i) and (ii), we get \[E=20.4\,V\] Note: It is evident from the equation \[V=E-ir\] larger the current \[i\] drawn from the cell, smaller will be the potential difference across its plates.
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