Electro Magnetic Induction And Alternating Currents Question 157
Question: If a direct current of value ampere is superimposed on an alternative current $ I=b,\sin ,\omega t $ flowing through a wire, what is the effective value of the resulting current in the circuit?
Options:
A) $ {{[ a^{2}-\frac{1}{2}b^{2} ]}^{1/2}} $
B) $ {{[ a^{2}+b^{2} ]}^{1/2}} $
C) $ {{[ \frac{a^{2}}{2}+b^{2} ]}^{1/2}} $
D) $ {{[ a^{2}+\frac{b^{2}}{2} ]}^{1/2}} $
Show Answer
Answer:
Correct Answer: D
Solution:
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as current any instant in the circuit will be
$ I=I_{dc}+I_{ac}=a+b,\sin ,\omega t $
so, $ I_{eff}={{[ \frac{\int_{0}^{T}{I^{2}dt}}{\int_{0}^{T}{dt}} ]}^{1/2}}={{[ \frac{1}{T}\int_{0}^{T}{{{(a+b,\sin ,\omega t)}^{2}}dt} ]}^{1/2}} $ i.e.
$ I_{eff}={{[ \frac{1}{T}\int_{0}^{T}{(a^{2}+2ab,\sin ,\omega t+b^{2},{{\sin }^{2}}\omega t)dt} ]}^{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}={{[ a^{2}+\frac{1}{2}b^{2} ]}^{1/2}} $
