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# GATE-2012 ECE Q10 (networks)

by on April 5, 2013

Question 10 on networks from GATE (Graduate Aptitude Test in Engineering) 2012 Electronics and Communication Engineering paper.

## Solution

The average power delivered to the load is integral of the instantaneous power over some period $T$, i.e.

$P_{avg}=\frac{1}{T}\int_TV_i(t)I_i(t)dt$

Given that $V_i(t)=I_i(t)R_L$, the equation reduces to,

$P_{avg}=\frac{1}{T}\int_TI^2_i(t)R_L dt=I^2_{rms}R_L$.

In our example with $I_i(t)={5\cos$$100\pi t + 100$$}$

$\begin{array}{lll}I^2_{rms}&=&\frac{1}{T}\int_TI^2_i(t)dt\\&=&\frac{1}{T}\int_T$$5\cos\(100\pi t + 100$$\)^2dt\\&=&\frac{5^2}{2}\end{array}$

Plugging in,

$P_{avg}=\|I_{rms}^\|^2{R_L}=\frac{25}{2}*4 = 50W$.

Based on the above, the right choice is (B) 50W

## References

[1] GATE Examination Question Papers [Previous Years] from Indian Institute of Technology, Madras http://gate.iitm.ac.in/gateqps/2012/ec.pdf

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