A) \[{{\left[ {{a}^{2}}-\frac{1}{2}{{b}^{2}} \right]}^{1/2}}\]
B) \[{{\left[ {{a}^{2}}+{{b}^{2}} \right]}^{1/2}}\]
C) \[{{\left[ \frac{{{a}^{2}}}{2}+{{b}^{2}} \right]}^{1/2}}\]
D) \[{{\left[ {{a}^{2}}+\frac{{{b}^{2}}}{2} \right]}^{1/2}}\]
Correct Answer: D
Solution :
[d] as current any instant in the circuit will be \[I={{I}_{dc}}+{{I}_{ac}}=a+b\,\sin \,\omega t\] so, \[{{I}_{eff}}={{\left[ \frac{\int_{0}^{T}{{{I}^{2}}dt}}{\int_{0}^{T}{dt}} \right]}^{1/2}}={{\left[ \frac{1}{T}\int_{0}^{T}{{{(a+b\,\sin \,\omega t)}^{2}}dt} \right]}^{1/2}}\] i.e. \[{{I}_{eff}}={{\left[ \frac{1}{T}\int_{0}^{T}{({{a}^{2}}+2ab\,\sin \,\omega t+{{b}^{2}}\,{{\sin }^{2}}\omega t)dt} \right]}^{1/2}}\] But as \[\frac{1}{T}\int_{0}^{T}{\sin \,\omega t\,dt=0}\] and \[\frac{1}{T}\int_{0}^{T}{{{\sin }^{2}}\,\omega t\,dt=\frac{1}{2}}\] So, \[{{I}_{eff}}={{\left[ {{a}^{2}}+\frac{1}{2}{{b}^{2}} \right]}^{1/2}}\]
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