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Chapter 1: 47E. (page 1)

Question: In a shunt-wound dc motor with the field coils and rotor connected in parallel (Fig. E27.47), the resistance \({R_f}\) of the field coils is \(106\;\Omega \), and the resistance \({R_r}\) of the rotor is \(5.9\;\Omega \). When a potential difference of \(120\;V\) is applied to the brushes and the motor is running at full speed delivering mechanical power, the current supplied to it is \(4.82\;A\).

(a) What is the current in the field coils?

(b) What is the current in the rotor?

(c) What is the induced emf developed by the motor?

(d) How much mechanical power is developed by this motor?

Short Answer

Expert verified

(a) The current through the field coil is \(1.13\;{\rm{A}}\).

Step by step solution

01

Given data

An applied voltage across the circuit:

\(V = 120\;{\rm{V}}\)

Current in the full circuit:

\(I = 4.82\;{\rm{A}}\)

Resistance of the field coil:

\({R_f} = 106\;\Omega \)

Resistance of the rotor coil:

\({R_r} = 5.9\;\Omega \)

02

Current through a resistance

The current through a resistance\(R\)when a potential difference\(V\)is applied across it is

\(I = \frac{V}{R}\) .....(i)

03

(a) Determination of the current through the field coil

Since the field coil and rotor are connected in parallel, potential difference across the field coil is\(V\). Thus, from equation (i), the current across it is

\(\begin{aligned}{c}I &= \frac{V}{{{R_f}}}\\ = \frac{{120\;{\rm{V}}}}{{106\;\Omega }}\\ = 1.13 \left( {1\;{{\rm{V}} \mathord{\left/{\vphantom {{\rm{V}} \Omega }} \right. \Omega } \times \frac{{1\;{\rm{A}}}}{{1\;{{\rm{V}} \mathord{\left/ {\vphantom {{\rm{V}} \Omega }} \right. \Omega }}}} \right)\\ = 1.13\;{\rm{A}}\end{aligned}\)

Thus, the required current is \(1.13\;{\rm{A}}\).

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