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The output of a step-down transformer is measured to be 24 V when connected to a 12 watt light bulb. The value of the peak current is |
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Answer» `1/sqrt2`A POWER associated with secondary `P_S`=12 W Now `P_S=I_S V_S` `therefore I_S=P_S/V_S=12/25`=0.5 A `therefore` Peak value of the current in the secondary `I_m=I_Sxxsqrt2` [ `because` Here, `(I_(rms))_S=I_S`] `therefore I_m=0.5xx1.414`=0.707 `therefore I_m=1/sqrt2A [ because 0.707 = 1/sqrt2]` |
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