Some generators, called shunt generators, use electromagnets...

Some generators, called shunt generators, use electromagnets in place of permanent magnets, with the field coils for the electromagnets activated by the induced voltage. The magnet coil is in parallel with the armature coil (it shunts the armature). As shown in Fig. $33-3$, a certain shunt generator has an armature resistance of $0.060 \Omega$ and a shunt resistance of $100 \Omega$. What power is developed in the armature when it delivers $40 \mathrm{~kW}$ at $250 \mathrm{~V}$ to an external circuit? From $\mathrm{P}=V I$ $$ \begin{array}{l} \text { Current to the external circuit }=I_{x}=\frac{\mathrm{P}}{V}=\frac{40000 \mathrm{~W}}{250 \mathrm{~V}}=160 \mathrm{~A} \\ \text { Field current }=I_{f}=\frac{V_{f}}{r_{f}}=\frac{250 \mathrm{~V}}{100 \Omega}=2.5 \mathrm{~A} \\ \text { Armature current }=I_{a}=I_{x}+I_{f}=162.5 \mathrm{~A} \end{array} $$ Total induced emf $=|\mathscr{E}|=\left(250 \mathrm{~V}+I_{a} r_{a}\right.$ drop in armature $$ \begin{aligned} &=250 \mathrm{~V}+(162.5 \mathrm{~A})(0.06 \Omega)=260 \mathrm{~V} \\ \text { Armature power } &=I_{a}|\mathscr{E}|=(162.5 \mathrm{~A})(260 \mathrm{~V})=42 \mathrm{~kW} \end{aligned} $$ $$ \begin{array}{l} \text { Alternative Method }\\ \begin{array}{l} \text { Power loss in the armature }=I_{a}^{2} r_{a}=(162.5 \mathrm{~A})^{2}(0.06 \Omega)=1.6 \mathrm{~kW} \\ \qquad \begin{aligned} \text { Power loss in the field } &=I_{f}^{2} r_{f}=(2.5 \mathrm{~A})^{2}(100 \Omega)=0.6 \mathrm{~kW} \\ \text { Power developed } &=(\text { Power delivered })+(\text { Power loss in armature })+(\text { Power loss in field) }\\ &=40 \mathrm{~kW}+1.6 \mathrm{~kW}+0.6 \mathrm{~kW}=42 \mathrm{~kW} \end{aligned} \end{array} \end{array} $$

Some generators, called shunt generators, use electromagnets in place of permanent magnets, with the field coils for the electromagnets activated by the induced voltage. The magnet coil is in parallel with the armature coil (it shunts the armature). As shown in Fig. $33-3$, a certain shunt generator has an armature resistance of $0.060 \Omega$ and a shunt resistance of $100 \Omega$. What power is developed in the armature when it delivers $40 \mathrm{~kW}$ at $250 \mathrm{~V}$ to an external circuit?
From $\mathrm{P}=V I$
$$
\begin{array}{l}
\text { Current to the external circuit }=I_{x}=\frac{\mathrm{P}}{V}=\frac{40000 \mathrm{~W}}{250 \mathrm{~V}}=160 \mathrm{~A} \\
\text { Field current }=I_{f}=\frac{V_{f}}{r_{f}}=\frac{250 \mathrm{~V}}{100 \Omega}=2.5 \mathrm{~A} \\
\text { Armature current }=I_{a}=I_{x}+I_{f}=162.5 \mathrm{~A}
\end{array}
$$
Total induced emf $=|\mathscr{E}|=\left(250 \mathrm{~V}+I_{a} r_{a}\right.$ drop in armature
$$
\begin{aligned}
&=250 \mathrm{~V}+(162.5 \mathrm{~A})(0.06 \Omega)=260 \mathrm{~V} \\
\text { Armature power } &=I_{a}|\mathscr{E}|=(162.5 \mathrm{~A})(260 \mathrm{~V})=42 \mathrm{~kW}
\end{aligned}
$$
$$
\begin{array}{l}
\text { Alternative Method }\\
\begin{array}{l}
\text { Power loss in the armature }=I_{a}^{2} r_{a}=(162.5 \mathrm{~A})^{2}(0.06 \Omega)=1.6 \mathrm{~kW} \\
\qquad \begin{aligned}
\text { Power loss in the field } &=I_{f}^{2} r_{f}=(2.5 \mathrm{~A})^{2}(100 \Omega)=0.6 \mathrm{~kW} \\
\text { Power developed } &=(\text { Power delivered })+(\text { Power loss in armature })+(\text { Power loss in field) }\\
&=40 \mathrm{~kW}+1.6 \mathrm{~kW}+0.6 \mathrm{~kW}=42 \mathrm{~kW}
\end{aligned}
\end{array}
\end{array}
$$

Key Concepts

Shunt Generator

A shunt generator is a type of DC machine where the field windings are connected in parallel (shunted) with the armature. This configuration allows the generator to automatically regulate its output voltage by feeding a portion of the generated voltage back into the field winding, influencing the strength of the magnetic field and hence the overall performance of the machine.

Induced Electromotive Force (EMF)

Induced emf is the voltage generated in the armature windings of a generator due to the change in magnetic flux, as described by Faraday's law of electromagnetic induction. In practical generator analysis, the total induced emf must account for the internal voltage drops due to resistances in the armature, providing a more complete picture of the actual generated voltage.

Armature Circuit Analysis

The armature circuit of a generator involves the current path through the armature windings, where the developed emf is reduced by the voltage drop across the armature resistance. Analyzing the armature circuit helps in determining the total current flow, and in calculating the losses within the generator caused by internal resistance effects.

Electrical Power Calculation

Calculating electrical power in circuits involves the basic relationship P = V × I, where power is the product of voltage and current. This formula is fundamental in assessing both the delivered power to an external load and the internal power developed within a generator, including the consideration of losses in various components.

Resistive Losses (I²R Losses)

Resistive losses, or I²R losses, refer to the power lost as heat when current flows through a resistance. In the context of generators, these losses occur in both the armature and field windings and must be accounted for when determining the overall efficiency and the developed power. Such losses are critical in practical electrical machines as they directly impact performance and energy conversion efficiency.

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