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A and B are two electric bulbs with their ratings respectively 40 W, 110 V and 100 W and 110 V. Find their respective filament resistances. If the bulbs are connected in series with a supply of 220 V, which bulb will fuse ? |
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Answer» Solution :For A bulb : `P_(1) = 40 Omega, V = 110` V Suppose resistance of bulb A is `R_(1)` . ` therefore P_(1) = (V^(2))/(R_(1))` `therefore R_(1) = (V^(2))/(P_(1))` `therefore R_(1) = (110 xx 110)/(40)= 302.5 Omega` So current flowing through it, `I_(1) = (V)/(R) = (110)/(302.5) = 0.3636 ` A For B bulb = `p_(1) = 1000 Omega,` V = 110 V Suppose resistance of bulb B is `R_(2)` , `therefore P_(2) = (V^(2))/(R_(2))` ` therefore R_(2)= (V^(2))/(P_(2))` ` therefore R_(2) = (110 xx 110)/( 100 ) = 121 Omega` `I_(2) = (V)/(R_(2)) = (110)/(121) = 0.9091 ` A Total resistance of circuit when both bulbs are 1V joined in series. `R= R_(1) + R_(2) = 302.5 + 121 = 423.5 Omega`  Current flowing through both bulbs I = `(V)/(R)` `= (220)/(423.5) = 0.52` A here , I` gt I_(1)` , So bulb A will be fused. I `LT I_(2) ` , so bulb B WILLNOT be damaged. |
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