question_answer

The cell has an emf of 2V and the internal resistance of this cell is 0.1\[\Omega \], it is connected to resistance of 3.9\[\Omega \] the voltage across the cell will be:

A)  1.95V                   

B)         1.5V                     

C)  2V                         

D)         1.8V

Correct Answer: A

Solution :

                Key Idea: When cell is giving current then the potential difference across its plates is less than its emf by potential drop across internal resistance.                                           When a cell of emf E is connected to a resistance of \[_{g}{{O}^{16}}\] then the emf E of the cell remains constant, while voltage V goes on decreasing on taking more and more current from the cell \[=16-8=8\]   \[_{2}H{{e}^{4}}\] where r is internal resistance. Also current \[=4-2=2\] \[_{26}F{{e}^{56}}\]   \[=56-26=30\] Putting the numerical values, we have \[_{92}{{U}^{235}}\] \[\text{m=}\frac{\text{image}\,\text{distance}}{\text{object}\,\text{distance}}\text{=-}\frac{\text{v}}{\text{u}}\text{=2}\] \[\Rightarrow \]