The circuit in figure shows two cells connected in opposition to each other. Cell \[{{E}_{1}}\] is of emf 6V and internal resistance \[2\,\Omega \] the cell is of emf 4V and internal resistances \[8\,\Omega .\] Find |
(i) effective resistance of the circuit |
(ii) effective emf of the two cells and current in the circuit |
(iii) the potential difference between the points A and B. |
Answer:
(i) Consider the given circuit as shown below, Effective resistance \[2\,\Omega +8\,\Omega =10\,\Omega \]
(ii) Effective emf of two cells \[=6-4=2\text{ }V,\] so the electric current is given by \[\Rightarrow \] \[I=2/10=0.2\,A\] along anti-clokwise direction, since \[{{E}_{1}}>{{E}_{2}}.\]
(iii) The direction of flow of current is always from high potential to low potential. Therefore, \[{{V}_{B}}>{{V}_{A}}.\] \[\Rightarrow \] \[{{V}_{B}}-4V-(0.2)\times 8={{V}_{A}}\] Therefore,
\[{{V}_{B}}-{{V}_{A}}=5.6V\]
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