The armature of a generator of resistance 1 , | vedclass

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(B) The electromotive force $(EMF)$ generated by the generator without load is $E = 125 \,V$.When the generator is under full load,the terminal voltag

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$10$

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(B) The electromotive force $(EMF)$ generated by the generator without load is $E = 125 \,V$.When the generator is under full load,the terminal voltage is $V = 115 \,V$.The internal resistance of the armature is $r = 1 \,\Omega$.The voltage drop across the internal resistance is given by $E - V = I \cdot r$.Substituting the given values: $125 \,V - 115 \,V = I \cdot 1 \,\Omega$.$10 \,V = I \cdot 1 \,\Omega$.Therefore,the current in the armature coil is $I = 10 \,A$.

The armature of a generator of resistance $1 \,\Omega$ is rotated at its rated speed and produces $125 \,V$ without load and $115 \,V$ with full load. The current in the armature coil is ......... $A$.

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