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A 50V, 10 W lamp is run on 100V, 50Hz ac mains Calculate the inductance of the choke coil required. |
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Answer» <P> Solution :Voltage marked on lamp, `V=50V`Power, P = 10W `therefore"RESISTANCE of lamp, R"=(V^(2))/(P)=(50xx50)/(10)=250Omega`. `"Current rating of lamp, "I_("rms")=(P)/(V)=(10)/(50)=(1)/(5)A` The given circuit is equivalent of LR circuit. Here `E_("rms")=100V, v=50Hz` When the lamp is worked on a.c., the impedance of the circuit is `Z_(L)=(E_("rms"))/(I_("rms"))=(100)/((1)/(5))="500 ohm"` `Z_(L)^(2)=R^(2)+Omega^(2)L^(2)(OR)Z_(L)^(2)=R^(2)+4pi^(2)v^(2)L^(2)` `therefore 4pi^(2)v^(2)L^(2)=z_(L)^(2)-R^(2) or L^(2)=(Z_(L)^(2)-R^(2))/(4pi^(2)^(2))` `L^(2)=((500)^(2)-(250)^(2))/(4XX(3.14)^(2)xx(50)^(2))=(187500)/(98596)=1.9 or L=1.38H` |
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