A DC shunt motor draws I_L = 50,text A from V = 220,text V . Given R_a = 0.4,Omega and R_f = 110,Omega, the back-EMF E_b and mechanical power developed are:

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Electrical Engineering QUESTION #2192

Question 1

A DC shunt motor draws $I_L = 50\,\text{A}$ from $V = 220\,\text{V}$. Given $R_a = 0.4\,\Omega$ and $R_f = 110\,\Omega$, the back-EMF $E_b$ and mechanical power developed are:

$E_b \approx 200.8\,\text{V}$, $P_{mech} \approx 9{,}638\,\text{W}$✔️
$E_b = 220\,\text{V}$, $P_{mech} = 11{,}000\,\text{W}$
$E_b = 196\,\text{V}$, $P_{mech} = 9{,}800\,\text{W}$
$E_b = 218\,\text{V}$, $P_{mech} = 10{,}900\,\text{W}$

Correct Answer Explanation

$I_f = V/R_f = 220/110 = 2\,\text{A}$. $I_a = I_L - I_f = 48\,\text{A}$. $E_b = V - I_a R_a = 220 - 48\times0.4 = 200.8\,\text{V}$. $P_{mech} = E_b\times I_a = 200.8\times48 \approx 9{,}638\,\text{W}$.

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