Shunt generator feeding six lamps

In a shunt generator the armature current supplies both the load (lamps) and the shunt field. The lamps are treated as fixed resistances found from their rating, and they now work at 220 V.

Given: $V = 220$ V, $R_a = 0.1\ \Omega$, $R_{sh} = 50\ \Omega$, six lamps of 200 V, 600 W, brush drop 1 V per brush.

Lamp current

Resistance of one lamp (from its rating):

$$R_{lamp} = \frac{V_{rated}^2}{P_{rated}} = \frac{200^2}{600} = 66.67\ \Omega$$

At 220 V, current through one lamp $= \dfrac{220}{66.67} = 3.3$ A.

Six lamps are in parallel, so

$$I_L = 6 \times 3.3 = 19.8\ \text{A}$$

Field and armature current

$$\begin{aligned} I_{sh} &= \frac{V}{R_{sh}} = \frac{220}{50} = 4.4\ \text{A} \\ I_a &= I_L + I_{sh} = 19.8 + 4.4 = 24.2\ \text{A} \end{aligned}$$

Generated emf

In a lap winding the brushes of the same polarity are in parallel, so the current path passes through one positive and one negative brush set. Total brush drop $= 2 \times 1 = 2$ V.

$$\begin{aligned} E_g &= V + I_a R_a + V_{brush} \\ &= 220 + 24.2 \times 0.1 + 2 \\ &= 220 + 2.42 + 2 = 224.42\ \text{V} \end{aligned}$$

   +----Ra---+------+--------+
   |         |      |        |
  (Eg)      Rsh   lamps x6   V=220 V
   |         |      |        |
   +---------+------+--------+

Answer: Armature current $I_a = 24.2$ A, generated emf $E_g = 224.42$ V.

(If the lamps were assumed to draw their rated current of 3 A each, $I_L = 18$ A, $I_a = 22.4$ A and $E_g = 224.24$ V. The resistance method above is more exact because the lamps are not at rated voltage.)