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A resistance of `4 Omega ` is connected across a cell Then it is replaced by another resistance of `1 Omega` it is found the power dissipeted in resistance in both the is `16` walt ThenA. internal resistance of the cell is `2 Omega`B. emf of the celll is `12V`C. maximum power the can be dissipated in the external resistance is `18` waltD. short circuit current from the cell is infinite |
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Answer» Correct Answer - a,b,c `((epsilon)/(4+r))^(2) xx 4 = ((epsilon)/(1+r))^(2) xx1` `r = 2 Omega` Current in circuit,`I = (epsilon)/(4+2)` `16 = ((epsilon)/(4+2))^(2) xx 4or epsilon = 12 V` power dissipated is maximum when exertnal resistance is equal to internal resistance i.e. `R = r = 2 Omega` `P_(max) = ((12)/(2+2))^(2) xx 2 = 18W` |
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