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### A sinusoidal voltage of peak value 250 V ....

A sinusoidal voltage of peak value 250 V is applied to a series LCR circuit, in which $R=8 \Omega$, $L=24 mH$ and $C=60 \mu F$. The value of power dissipated at resonant condition is 'x' kW. The value of x to the nearest integer is _ _ _ _ .

Solution

At resonance condition, the full voltage is applied across the resistor.

So, power dissipated $= \frac {V^2 _{rms}}{R} = \frac {(\frac {250}{\sqrt 2 })^2} {8} = 3906.25 W \approx 3.91 kW$

$\therefore x = 4$

### A man starts walking from the point P (-3, 4) ....

A man starts walking from the point P (-3, 4), touches the x-axis at R, and then turns to reach at the point Q (0, 2). The man is walking at a constant speed. If the man reaches the point Q in the minimum time, then $50 [(PR)^2 + (RQ)^2 ]$ is equal to _ _ _ _ . Solution For time to be minimum at constant speed, the directions must be symmetric. In other words, the angles made by PR and RQ with the vertical must be the same just like in the law of reflection in optics. $tan \theta = \frac {MP}{MR} = \frac {NQ}{NR}$ $\Rightarrow \frac {3-r}{4} = \frac {r}{2}$ $\Rightarrow r=1$ So, $R \equiv ( - 1,0)$ Now, $50(PR^2+RQ^2)=50[(4+16)+(1+4)]=1250$