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## Q. 5.11

DC Motor

A motor, when switched on, does not immediately attain its final speed. This is due to the inductance of the motor coils. A DC motor is connected as shown in Figure 5.30 (a) and its model is shown in Figure 5.30 (b). Assume an inductor with zero initial charge (or assume the switch is open for a long time period). The motor reaches its steady-state speed when $v_{\text {motor }}(t)=V_{S}$. DC and AC motors have been discussed in detail in Chapter 13.

a. Find $v_{\text {motor }}\left(0^{+}\right)$and $v_{\text {motor }}(\infty)$

b. Find an expression for $v_{\text {motor }}(t)$

c. Find the time at which $v_{\text {motor }}(t)=95 \%$ of $V_{S}\left(V_{S}=120 V\right)$ ## Verified Solution

a. The inductor is an open circuit at t = 0; thus, $v_{\text {motor }}(0)=0 V$. The inductor is a short circuit at t = ∞ ; thus, $v_{\text {motor }}(\infty)=120 V$.

b. Applying KVL for t > 0:

$120-v_{L}(t)-v_{\text {motor }}(t)=0$

$v_{\text {motor }}(t)=10 i_{L}(t)$

and:

$v_{L}(t)=0.13 \frac{di_{L}(t)}{dt}$

Thus:

$120-0.13 \frac{di_{L}(t)}{dt}-10 i_{L}(t)=0$

Solving for $i_{L}(t)$ yields:

$i_{L}(t)=12\left(1-e^{\frac{-t}{1.3 \times 10^{-2}}}\right)$

Accordingly:

\begin{aligned}v_{\text {motor }}(t) &=10 \times 12\left(1-e^{\frac{-t}{13 \times 10^{-2}}}\right) \\&=120\left(1-e^{\frac{-t}{13 \times 10^{-2}}}\right)\end{aligned}

c. $v_{\text {motor }}(t)=0.95 V_S$, thus:

$120 \times 0.95=120\left(1-e^{\frac{-1}{13 \times 10^{-2}}}\right)$

Accordingly:

$t=-1.3 \times 10^{-2} \times \ln 0.05=38.95 ms$