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PTCUL AE E&M 2017 Official Paper (Set B)

Option 1 : 4.8 Ω

**Concept of Armature Rheostatic Speed Control:**

This method is used when speeds below the no-load speed are required. As the supply voltage is normally constant, the voltage across the armature is varied by inserting a variable rheostat or resistance (called controller resistance) in series with the armature circuit as shown in Fig.

Let,

I_{a1} = armature current in the first case

I_{a2} = armature current in the second case

(If I_{a1} = I_{a2}, then the load is of constant torque.)

N_{1}, N_{2} = corresponding speeds, V = supply voltage

then N_{1} ∝ V − I_{a1}R_{a} ∝ E_{b1}

Where ,E_{b1} = V - I_{a1}R_{a}

Let some controller resistance of value R be added to the armature circuit resistance so that its value becomes,

(R + R_{a}) = R_{t}

Then,

N_{2} ∝ V − I_{a2} R_{t} ∝ E_{b2}

\(=\frac{{{N_2}}}{{{N_1}}} = \frac{{{E_2}}}{{{E_1}}}\)

From the above equation, we get the value of E_{b2},

To calculate External resistance value,

\({E_{b2}} = {E_{b1}} - {I_a}\left( {R + {R_a}} \right)\;V\)

**Calculation:**

**Given:**

V = 220 V, N_{1} = 1000 rpm, I_{a} = 17.5 A, N_{2} = 600rpm, R_{a} = 0.4 ohm

Eb1 = V - Ia1Ra

= 220 - 17.5 × 0.4

= 213 V

With constant flux,

E_{b} ∝ N

\(=\frac{{{N_2}}}{{{N_1}}} = \frac{{{E_2}}}{{{E_1}}}\)

\({E_{b2}} = \frac{{600}}{{1000}} \times 213 = 127.8\;V\)

\({E_{b2}} = {E_{b1}} - {I_a}\left( {R + {R_a}} \right)\;V\)

\(220 - \left( {R + 0.4} \right)17.5 = 127.8\)

\(\left( {R + 0.4} \right) = \frac{{220 - 127.8}}{{17.5}}\)

**R ≈ 4.8 Ω**