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In the circuit givenn below, both batteries are ideal. EMF E_(1) of battery 1 has a fixed value but emf E_(2) of battery 2 can be varied between 1.0 V and 10.0 V the graph gives the currents through the two batteries as a function of E_(2), but are not marked as shown plot corresponds to which battery. but both plots, current is assumed to be negative when the direction of the current through the battery is opposite the direction of that battery's emf. (direction of emf is from negative to positive). Q. The resistance R_(1) has value |
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Answer» `10Omega`   `i_(1)=0.1A,E_(2)=4V` `i_(2)=0` As `0.1R_(1)+0.1R_(2)+0.1R_(2)-E_(1)=0.1` `R_(2)-4V=0` `R_(2)=40Omega` Now `i_(2)=0.3A`, `i_(1)=-0.1A` `E_(2)=8V` Now `0.1R_(1)+E_(1)-8=0` `0.1+3-E_(1)=0` `0.2R_(1)-4=0impliesR_(1)=(4)/(0.2)=20Omega` `E_(1)=2+4=6V` |
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