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Haryana PMT Haryana PMT Solved Paper-1999

  • question_answer
    Four cells each of en-if 1.5V and internal resistance 2\[\Omega \] are connected in parallel. This combination sends a current to an external resistance 2\[\Omega \]. The value of current in external resistance is :

    A)  1.8 A                                    

    B)  1.5 A

    C)  0.8 A                                    

    D)  0.6 A

    Correct Answer: D

    Solution :

                    Net emf of combination of cells in parallel order = emf of each cell \[=1.5V\] Net internal resistance is                 \[{{r}_{in}}=\frac{r}{n}=\frac{2}{4}=0.5\Omega \] external resistance \[R=2\Omega \] So, current is given by                 \[i=\frac{E}{R+{{r}_{in}}}=\frac{1.5}{2+0.5}\]                 \[=\frac{1.5}{2.5}=0.6A\]


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