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A network of resistors is connected to a 12V battery as shown in fig. a) Calculate the equivalent resistance of the network. b) Obtain current in 12Omega and 6Omega resistors. |
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Answer» SOLUTION :`R_(1)=12OMEGA` `R_(2)=6Omega` `V=12V` `R_(P)=?` `I_(1)=?` `I_(2)=?` The EQUIVALENT resistance of the parallel combination is `(1)/(R_(p))=(1)/(R_(1))+(1)/(R_(2))` `R_(p)=(R_(1)R_(2))/(R_(1)+R_(2))=(12xx6)/(12+6)` `R_(p)=4Omega` The current through resistor `R_(1)=12Omega` is `V=I_(1)R_(1)` `I_(1)=(V)/(R_(1))=(12)/(12)=1A` The current through resistor `R_(2)=6Omega` is `V=I_(2)R_(2)` `I_(2)=(V)/(R_(2))=(12)/(6)=2A` |
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