The open-loop transfer function of a dc motor is given as $\frac{ω (s)}{V_{a} (s)} = \frac{10}{1 + 10 s}$ . When connected in feedback as shown below, the approximate value $K_{a}$ that will reduce the time constant of the closed loop system by one hundred times as compared to that of the open-loop system is

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Solution

The correct option is C 10
$\frac{ω (s)}{V_{a} (s)} = \frac{10}{1 + 10 s}$
$1 + T s = 1 + 10 s$
$T s = 10 s$
$T = 10$
$T^{'} = \frac{T}{100} = \frac{100}{100} = \frac{1}{10}$
$\frac{ω (s)}{R (s)} = \frac{K_{a} \times \frac{10}{1 + 10 s}}{1 + K_{a} \frac{10}{1 + 10 s}}$

$\frac{ω (s)}{R (s)} = \frac{10 K_{a}}{(1 + 10 s + 10 K_{a})}$

$= \frac{10 K_{a}}{(1 + 10 K_{a}) + 10 s}$

$\frac{ω (s)}{R (s)} = \frac{10 K_{a}}{(1 + 10 K_{a}) [1 + \frac{10 s}{1 + 10 K_{a}}]}$

$T^{'} = \frac{10}{1 + 10 K_{a}} = \frac{1}{10}$
$1 + 10 K_{a} = 100$
$10 K_{a} = 99$
$K_{a} = 9.9 = 10$

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The open-loop transfer function of a dc motor is given as ω(s)Va(s)=101+10s. When connected in feedback as shown below, the approximate value Ka that will reduce the time constant of the closed loop system by one hundred times as compared to that of the open-loop system is

The correct option is C 10ω(s)Va(s)=101+10s 1+Ts=1+10s Ts=10s T=10 T′=T100=100100=110 ω(s)R(s)=Ka×101+10s1+Ka101+10s ω(s)R(s)=10Ka(1+10s+10Ka) =10Ka(1+10Ka)+10s ω(s)R(s)=10Ka(1+10Ka)[1+10s1+10Ka] T′=101+10Ka=110 1+10Ka=100 10Ka=99 Ka=9.9=10