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Question-18460




Question Number 18460 by tawa tawa last updated on 21/Jul/17
Answered by sandy_suhendra last updated on 23/Jul/17
P_(bulb) =5 w  V_(bulb) =170 V  I_(bulb)  = (P_(bulb) /V_(bulb) ) = (5/(170)) Ampere = I_(resistance)   V_(resistance)  = 380−170 = 210 V  R_(resistance)  = (V_(res) /I_(res) ) = ((210)/(5/170)) = 7140 ohm (C)
$$\mathrm{P}_{\mathrm{bulb}} =\mathrm{5}\:\mathrm{w} \\ $$$$\mathrm{V}_{\mathrm{bulb}} =\mathrm{170}\:\mathrm{V} \\ $$$$\mathrm{I}_{\mathrm{bulb}} \:=\:\frac{\mathrm{P}_{\mathrm{bulb}} }{\mathrm{V}_{\mathrm{bulb}} }\:=\:\frac{\mathrm{5}}{\mathrm{170}}\:\mathrm{Ampere}\:=\:\mathrm{I}_{\mathrm{resistance}} \\ $$$$\mathrm{V}_{\mathrm{resistance}} \:=\:\mathrm{380}−\mathrm{170}\:=\:\mathrm{210}\:\mathrm{V} \\ $$$$\mathrm{R}_{\mathrm{resistance}} \:=\:\frac{\mathrm{V}_{\mathrm{res}} }{\mathrm{I}_{\mathrm{res}} }\:=\:\frac{\mathrm{210}}{\mathrm{5}/\mathrm{170}}\:=\:\mathrm{7140}\:\mathrm{ohm}\:\left(\mathrm{C}\right) \\ $$

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