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In the given circuit values are as follows epsilon_(1)=2V, epsilon_(2)=4V, R_(1)=1Omega and R_(2)=R_(3)=1Omega. Calcualte the Currents through R_(1), R_(2) and R_(3). |
Answer» Solution : LET `i_(1), i_(2)`are CURRENTS across `R_(1) and R_(3)`. `(i_(1)+i_(2))` is current across `R_(2)`. Their DIRECTION are taken as shown From Kirchoff.s second LAW for AGFBA LOOP `-i_(1)R_(1)-(i_(1)+i_(2))R_(2)+E_(1)=0` `i_(1)+i_(1)+i_(2)=2 rArr 2i_(1)+i_(2)=2rarr(1)` From Kirchooff.s second law for BCDEB loop `-i_(2)R_(3)-(i_(1)+i_(2))R_(2)+E_(2)=0` `i_(2)+i_(1)+i_(2)=rArr i_(2)+2i_(2)=4rarr(2)` Solving equation (1) and (2) we get `i_(1)=0A, i_(2)=2A` Thus currents across `R_(1)` is 0, while across `R_(3)` and R_(2)` are 2A each |
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