By Carl E. Smith

Arithmetic, consultant, How-to, Communications, Engineering

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**Extra resources for Applied mathematics for radio and communication engineers**

**Example text**

Proceeding as before, using Eq. 5) = -10 = 20( - VR; R; Note: The square of a negative number is a positive number. 22 Ans. The angle of the resultant vector R by Eq. (12) is R. 485 Note: This is negative because it is in the second quadrant instead of the first. From trigonometric tables, 'Y = 82°56' This, however, is the angle vector R makes with the negative x-axis. If we subtract this from 180° we get the angle" vector R makes with the positive It will be noted that this equation is like Eq.

__ + X2 R2]2 1 2 - X2 + R2 Squaring both sides (Rule 7) and turning the equation around, R2 yE2 - = Transposing J2 R2, Dividing by (X2 Dividing by I, y- I Squaring both sides to remove the radical, E Factoring 1 2 , 1=- yR2 + X2 E - yE2 - R2I2 Clearing of fractions, Solving for R was slightly more complicated but each step in itself was quite simple. The student should write out each step when solving problems until the procedure is thoroughly under stood. If the steps are written out in detail, mistakes will be less likely to occur.

This chflcks the value given above. = 60 10 = - ylO,OOO - 6,400 10 = 6 ohms reactance 52 ALGEBRA APPLIED MATHEMATICS In case the student may not know when to square or when to extract a square root, it may be helpful to remember that X = (VX)2, so if X2 is given, extract the square root to get X and if VX is given, square the quantity to get X. Of COurse rules 7 and 8 must be remembered when performing these operations on equations. Both sides of the equation must be treated alike. In the equation, XI = Xc we know that XI = 27rfL and 1 Xc = 27rfC Therefore, by substituting as pointed out in rule 10, we can write 27rfL = 1 27rfC Exercises 1.

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