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Question Number 16835 by tawa tawa last updated on 26/Jun/17

Answered by sandy_suhendra last updated on 27/Jun/17

1)  5C3×5C3×4C2  = 10×10×6  = 600 ways  2) V_(out)  = 15 v  V_(in ) = 220 v  I_(in)  = 0.7 Ae  a) P_(in)  = V_(in) .I_(in)  = 220×0.7 = 154 watt  P_(out)  = 3×40 = 120 watt  η = (P_(out) /P_(in) ) × 100% = ((120)/(154))×100% = 77.9%  b) the cost = ((154×24)/(1000)) × 30k = 110.88k

$$\left.\mathrm{1}\right) \\ $$$$\mathrm{5C3}×\mathrm{5C3}×\mathrm{4C2} \\ $$$$=\:\mathrm{10}×\mathrm{10}×\mathrm{6} \\ $$$$=\:\mathrm{600}\:\mathrm{ways} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{V}_{\mathrm{out}} \:=\:\mathrm{15}\:\mathrm{v} \\ $$$$\mathrm{V}_{\mathrm{in}\:} =\:\mathrm{220}\:\mathrm{v} \\ $$$$\mathrm{I}_{\mathrm{in}} \:=\:\mathrm{0}.\mathrm{7}\:\mathrm{Ae} \\ $$$$\left.\mathrm{a}\right)\:\mathrm{P}_{\mathrm{in}} \:=\:\mathrm{V}_{\mathrm{in}} .\mathrm{I}_{\mathrm{in}} \:=\:\mathrm{220}×\mathrm{0}.\mathrm{7}\:=\:\mathrm{154}\:\mathrm{watt} \\ $$$$\mathrm{P}_{\mathrm{out}} \:=\:\mathrm{3}×\mathrm{40}\:=\:\mathrm{120}\:\mathrm{watt} \\ $$$$\eta\:=\:\frac{\mathrm{P}_{\mathrm{out}} }{\mathrm{P}_{\mathrm{in}} }\:×\:\mathrm{100\%}\:=\:\frac{\mathrm{120}}{\mathrm{154}}×\mathrm{100\%}\:=\:\mathrm{77}.\mathrm{9\%} \\ $$$$\left.\mathrm{b}\right)\:\mathrm{the}\:\mathrm{cost}\:=\:\frac{\mathrm{154}×\mathrm{24}}{\mathrm{1000}}\:×\:\mathrm{30k}\:=\:\mathrm{110}.\mathrm{88k} \\ $$$$ \\ $$

Commented by tawa tawa last updated on 27/Jun/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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