একটি 250V DC shunt মোটরের আর্মেচারের রোধ 0.25 $\Omega$  আছে কারেন্ট 50Amp ও speed 750rpm । যদি লোড টর্কের পরিচ ছাড়াই মোটরের ফিল্ড ফ্লাক্স 10% কমে যায় তাহলে মোটরে নতুন কত হবে?

A 250V DC shunt motor has an armature resistance of 0.25 $\Omega$ , a current of 50Amp and a speed of 750 rpm. If the field flux of the motor is reduced by 10% without any change in load torque, what will be the new torque in the motor?

Maximum Power Transfer Theorem :

In electrical engineering, the maximum power transfer theorem states that, to obtain maximum external power from a power source with internal resistance, the resistance of the load must equal the resistance of the source as viewed from its output terminals.

For t<0, the switch is closed.

Capacitor acts as open to DC.

Voltage across the capacitor,

     v(0^-)= {(12||4)×V}÷{(12||4)+6} = 8VFor t=0, the switch is opened,Voltage across the Capacitor cannot change instantaneously,So, v(0^-)= v(0) = 8VAt t>0, The capacitor is discharging.R_th = (12||4) = 3 ohmC= 1/6 FTime constant, Π = R_th×C = 0.5 So, v(t) = v(0)e^-t/Π  = 8e^-t/0.5 =8e^-2t V (Ans.)