A) \[\sqrt{\left[ {{a}^{2}}+\frac{{{b}^{2}}}{2} \right]}\]
B) \[\sqrt{\left[ {{a}^{2}}+\frac{{{b}^{2}}}{4} \right]}\]
C) \[\sqrt{\left[ a+\frac{b}{2} \right]}\]
D) \[\sqrt{\left[ a+\frac{b}{4} \right]}\]
Correct Answer: A
Solution :
[a] : The current in the circuit at any instant is, \[I=a+b\sin \omega t.\] Hence, the effective value, \[{{I}_{eff}},\]of the current in the circuit will be given by \[I_{eff}^{2}=\left[ \frac{\int\limits_{0}^{T}{{{I}^{2}}dt}}{\int\limits_{0}^{T}{dt}} \right]=\frac{1}{T}\int\limits_{0}^{T}{{{I}^{2}}dt}=\frac{1}{T}\int\limits_{0}^{T}{{{[a+bsin\omega t]}^{2}}dt}\] \[\Rightarrow \]\[I_{eff}^{2}=\frac{1}{T}\int\limits_{0}^{T}{[{{a}^{2}}+2absin\omega t+{{b}^{2}}si{{n}^{2}}\omega t]dt}\],(i) Now, we have \[\frac{1}{T}\int\limits_{0}^{T}{sin\omega dt=0and\frac{1}{T}\int\limits_{0}^{T}{si{{n}^{2}}\omega t}\,dt=\frac{1}{2}}\] Substituting these values in equation (i), we get \[{{I}_{eff}}=\sqrt{\left[ {{a}^{2}}+\frac{{{b}^{2}}}{2} \right]}\]
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