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When a resistance of 11 Omega is connected in series with an electric cell, 0.5A current flows thrugh it . If the 11 Omega resistor is replaced by 5 Omega resistor the current flowing through it is 0.9 A . Find the internal resistance of the cell. |
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Answer» Solution :When the 11 `OMEGA` resistance is CONNECTED in series. i= 0.5 A, R = 11 Omega` `V=i(R+r)` = where is the INTERNAL in series. `=0.5 (11+r)""….(1)` When the `5Omega` resistance is connected in series i=0.9 A `V= 0.9 (5+r)"".....(2)` From (1)& (2) , `0.5(11 +r) = 0.9(5+r)` `0.5xx11+0.5 r = 0.9xx5+0.9r` `5.5 + 0.5r=4.5 + 0.9r ` `0.4r=1` `r=2.5 Omega` |
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