Solutions to Problems

Frontmatter

Chapter 1

Abstract

The equivalent circuit is shown in figure Alla, which reduces to that of figure Al.lb since the 8 Ω and 1 kΩ are in parallel and by the product-over-sum rule these combine to give a resistance of

Then the total series resistance is 2 + 0.1 + 7.9365 = 10.0365 Ω and the current, I, is 9/10.0365 = 0.8967 A. Then the voltage across the 7.9365Ω resistance is 0.8967 × 7.9365 = 7.117 V, which is the required voltmeter reading.

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Chapter 2

Abstract

Consider figure A2.la, in which the waveform is triangular and symmetrical about the t-axis, so we need only consider the positive part.

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Chapter 3

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Chapter 4

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Chapter 5

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Chapter 6

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Chapter 7

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Chapter 8

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Chapter 9

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Chapter 10

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Chapter 11

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Chapter 12

Abstract

1 Consider a small element of the circumference of the coil, as shown in figure Al2.1, whose length is δl and which is inclined at an angle, ø, to the vertical. The Lorentz force on this element is

where δI is directed along the tangent to the circumference in the direction of i. The force’s direction is given by the right-hand corkscrew rule. When the plane of the coil makes no angle (θ = 0°) with B the angle between δ1 and B is (90° — ø) and the force’s magnitude is

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Chapter 13

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Chapter 14

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Chapter 15

Abstract

1 Using the e.m.f. equation for the side with the most turns (the secondary) we find

Whence

And the primary turns are then

The nearest whole number is 61.

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Chapter 16

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Chapter 17

Abstract

1 Figure A17.1a shows the equivalent circuit of one phase of the star-connected generator, in which the line voltage of 6.9 kV has been converted to the phase voltage of 6.9/⇃ 3 kV. Attached to this phase of the generator is an 8Ω load (R_LS), which is the star-equivalent of the 24Ω delta-connected load.

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Chapter 18

Abstract

1 The power consumption is

The diodes are conducting a third of the times each, so that at any one time there are two conducting, with a combined voltage drop of 1.4 V. The current flowing in the diodes is the load current, I_DC which is E_DC /R or 1.35 × 415/9.5 = 58.97 A, and then the power consumed by the diodes is 1.4 × 58.97 = 82.6 W. As a fraction of the power output thus is 82.6/33000 = 0.0025 or 0.25%, a negligible amount.

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Chapter 19

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Chapter 20

Abstract

1 We note that at all times D₂ = Q₁ and that D₁ = (Q₁Q₂)’ and that the device is positive-edge triggered. The table of figure A20.1a gives the successive states of the circuit when the initial state is Q₁ = Q₂ = 1, and the corresponding timing diagram is shown in figure A20.1b. When the initial state is Q₁ = Q₂ = 0, the successive states are shown in figure A20.1c, from which we see that the former sequence repeats.

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Chapter 23

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Chapter 24

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Chapter 25

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Chapter 26

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Chapter 27

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Chapter 28

Abstract

1 The voltage 6.31 ± 0.3% can be written 6.31 ± 0.02 V, so that adding it to 3.63 ± 0.03 V produces 9.94 ± 0.05 V, making the percentage error 0.05 × 100/9.94 = 0.5%. This is a conservative way of estimating the error.

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