Question

A DC shunt motor takes an rated armature current of 50 A at its rated voltage of 240 V. Its armature circuit resistance is 0.2 $\Omega$. If an external resistance of 1 $\Omega$ is inserted in series with the armature and field flux remains unchanged. The percentage increase or decrease in speed for half of the load torque is

• A. 8.7% increase
• B. 21. 739% decrease
• C. 8.7% decrease
• D. 21.739% increase

Answer 1

The correct option is C 8.7% decrease

$I_a=50A,$ $V_t=240V$
$R_a=0.2\Omega$
$E_1=V_t-I_a{R}_a$
$=240-50\times 0.2$
$=230V$
$\text{Now as torque is reduced to half the rated load torque}$
$T\propto \varphi I_a$
$\text{as flux is constant}$
$T\propto I_a$

So, $\frac{T_1}{T_2}=\frac{I_{a1}}{I_{a2}}$

$\frac{1}{0.5}=\frac{50}{I_{a2}}$

$I_{a2}=25A$

$\text{1}\Omega \text{resistor is added to armature circuit as external resistor}$
$E_2=V_t-I_{a2}(R_a+R_{ext})$
$E_2=240-25\left(1.2\right)$
$E_2=210V$
$E\propto \varphi N$
$\text{as flux is constant,}$ $E\propto N$

$\frac{E_1}{E_2}=\frac{N_1}{N_2}$

$\frac{230}{210}=\frac{N_1}{N_2}$

$N_2=0.913N_1$

$\text{\% change in speed}=\frac{N_2-N_1}{N_1}\times 100\%=\frac{0.913N_1-N_1}{N_1}\times 100\%=8.7\%decrease$