InterviewSolution
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InterviewSolution
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Two heaters designed for the same voltage `V` have different power ratings. When connected individually across as source of voltage` V`, they produce `H` amount of heat each in time `t_1` and `t_2` respectively. When used together acros the same source, they produce H amount of heat in time tA. if they are in series `t=t_1+t_2`B. if they are in series `t=2(t_1+t_2)`C. if they are in parallel `t=(t_1t_2)/((t_1+t_2))`D. if they are in parallel `t=(t_1t_2)/(2(t_1+t_2))` |
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Answer» Correct Answer - A::C `H=V^2/R_1 t_1implies R_1=(V^2t_1)/H` Similarly, `R_2=(V^2t_2)/H` In series, `H=(V^2/(R_1+R_2))t` `t=(H(R_1+R_2))/V^2` substituting the value of `R_1` and `R_2` we get `t=t_1+t_2` In parallel, `H=V^2/R_("net")t=V^2t(1/R_1+1/R_2)` `=V^2t(H/(V^2t_1)+H/(V^2t_2))` Solving we get, `t=(t_1t_2)/(t_1+t_2)` |
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