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An LR circuit constainsan inductor of 500 mH,a resistor of 25.O Omega and an emf of 5.00 Vin series. Find the potential difference across the resistor at t =(a) 20.0 ms, (b) 100 ms and ( c) 1.00 s. |
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Answer» SOLUTION :Here `L=500mH, R=25 Omega, E=5V` `(a) For t=20 ms` ` i=i_0(1-e^(-tR//L)` `=(E/R)(1-e^(-tR//L))` `=(5/25)(1-e^(-(2xx10^(-3)xx25)//(500xx10^(-3))` `=1/5(1-e^(-1)=(1/5)(1-0.3678)` `=(0.632)/(5)=0.1264`.` `POTENTIAL difference ` `=iR=C.1264xx25` `=3.1606V=3.16V` `(b) Here t=100ms` `i-i_0(1-e^(-tR//L))` `=(5/25)(1-r^(-100)xx10^(-3)xx25//500xx10^(-3))` `=1/5(1-e^(-50)` =(1/5)(1-0.0067)` `=(0.9932)/5=0.19864` potential difference iR `0.19864xx25xx4.9665=4.97 V.` ` (c ) Here t=1 SEC.` `i=(5/25)(1-e^(-1xx(25/100)xx10^(-3))` `=(1/5)(1-e^(-50))` `=1/5xx1=(1/5)A` potential difference ` `=(1/5 xx25)V=5V.` |
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