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Two resistors are connected in series with 5V battery of negligible internal resistance. A current of 2 A flows through each resistor. If they are connected in parallel with the samebattery a current of (25)/(3)A flows through combination. Calculate the value of each resistance. |
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Answer» <P> Solution :Resistors in series `: R _(S) = R_(1) + R _(2)``I = (epsi )/(R _(s))` `R _(S) = (epsi )/(I) =5/2` `R_(1) + R _(2) =2.5` Resistors in PARALLEL `R _(p) = (R _(1) R_(2))/( R _(1) + R _(2))` `I = (epsi )/(R _(p))` `R _(p) = (epsi )/(I) = (5)/( 25//3) = 3/5` `R_(1) R _(2) = 3/5 XX (R_(1) + R _(2)) = 3/5 xx 2.5` `R_(1) R _(2) = 1.5` Solving equations (1) and (2),` R_(1) = 1.5 Omega and R_(2) = 1OMEGA ` |
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