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An ideal battery passes a current of 5A through a resistor. When it is connected to another resistance of 10Omega in parallel, the current is 6A. Find the resistance of the first resistor. |
Answer» Solution : `"Current through "R_(1)" in the FIRST case "i_(1)=5A` `"Current in the second case "i_(2)=6A` EFFECTIVE resistance in the second case `R=(R_(1)R_(2))/(R_(1)+R_(2)), V=I_(1)R_(1) and V=I_(2)(R_(1)R_(2))/(R_(1)+R_(2))` `I_(1)R_(1)=I_(2)(R_(1)R_(2))/(R_(1)+R_(2))RARR I_(1)=I_(2)(R_(2))/(R_(1)+R_(2))` `5=6xx(10)/(R_(1)+10)rArr 5(R_(1)+10)=60` `5R_(1)+50=60, 5R_(1)=10` `R_(1)=(10)/(5)=2OmegarArr R_(1)=2Omega` |
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