## Chapter 5

## 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)

## Step-by-Step

## 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)=0In addition:

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)=0Solving 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