No load V_{t}=6 V , I_{ mo }=14.5 mA , n=12125 rpm or \omega=1269.7 rad / s
(a) E_{a}=6-14.5 \times 10^{3} \times 4.2= 5.939 V
5.939=K_{m} \omega=K_{m} \times 1269.7
Or K_{m}=4.677 \times 10^{-3}
(b) Rotational loss, P_{r o t}=E_{a} I_{a} ; there is no load
=5.939 \times 14.5 \times 10^{-3}=0.0861 W
(c) Stalled current
\omega=0 so E_{a}=0
I_{a}( stall )=\frac{6}{4.2}=1.4285 A
Torque (stall) =K_{m} I_{a} \text { (stall) }=4.677 \times 10^{-3} \times 1.428=6.67 m Nm
(d) P_{\text {out }}(\text { gross })=1.6 W =E_{a} I_{a}
\left(6-4.2 I_{a}\right) I_{a}=1.6
4.2 I_{a}^{2}-6 I_{a}+1.6=0
Solving we find I_{a}=0.354 A , 1.074 A
Thus
I_{a}=0.354 A ; higher value rejected
E_{a}=6-0.854 \times 4.2=4.513 V =K_{m} \omega
\omega=\frac{4.513 \times 10^{3}}{4.677}=965 rad / s
Rotational loss (proportional to square of speed)
P_{r o t}=0.0861 \times\left(\frac{965}{1269.7}\right)^{2}= 0.05 W
P_{\text {out }}(\text { net })=P_{\text {out }}(\text { gross })-P_{\text {rot }}=1.6-0.05= 1.55 W
Power input, P_{i}=V_{t} I_{a}=6 \times 0.354= 2121 W
\eta=\frac{1.55}{2.124} \times 100=73 \%
(e) Motor speed, n = 10250 rpm or \omega=1073.4 rad / s
E_{a}=K_{m}^{\omega}=4.513 \times 10^{-3} \times 1073.4= 4.844 V
I_{a}=\frac{6-4.844}{4.2}= 0.275 A
P_{\text {out }}(\text { gross })=P_{e}=E_{a} I_{a}=4.844 \times 0.275= 1.332 W
P_{r o t}=0.0861 \times\left(\frac{1073.4}{1269.7}\right)^{2}= 0.0615 W
P_{o u t}(\text { net })=1.332-0.0615= 1.27 W
P_{i n}=6 \times 0.275= 1.65 W
\eta=\frac{1.27}{1.65} \times 100=77 \%