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Alicia C.
Electrical Engineering Major at Clemson University
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Microsoft Excel
TutorMe
Question:

Find the sum of E13 to E22 and A4 to F4.

Alicia C.
Answer:

=SUM(E13:E22)+SUM(A4:F4)

Calculus
TutorMe
Question:

What is the derivative of $${2x^2+7x}\over{5x^4}$$

Alicia C.
Answer:

The quotient rule I like to think as "low dehigh minus high delow over low squared" as in the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator over the denominator squared. $${d}\over{dx}$$$${2x^2+7x}\over{5x^4}$$=$${5x^4(4x+7)-(2x^2+7x)(20x^3)}\over{25x^8}$$=$${x(4x+7)-(2x^2+7x)(4)}\over{5x^5}$$=$${4x^2+7x-8x^2-28x}\over{5x^5}$$=$${-4x^2-21x}\over{5x^5}$$=$${-4x-21}\over{5x^4}$$

Electrical Engineering
TutorMe
Question:

A parallel circuit has three resistors, R1, R2, R3, and a voltage source of 50 volts. R1 is in series with the voltage source and R2 and R3 are in parallel with each other. If R1=4 ohms, R2=20 ohms, and R3=80 ohms, then: a) what is the current through each resistor, b) what is the voltage through each resistor, and c) verify the total power developed equals the total power absorbed.

Alicia C.
Answer:

First, find the total resistance, RT, of the circuit. RT=R1+(R2*R3)/(R2+R3)=20 ohms. Using RT and Vs, we can find the total current, IT, which is equal to the current through R1, I1. IT=I1=Vs/RT=2.5A. With this we can find the voltage through R1. V1=R1*I1=10V. Now find the voltage through the R2 and R3 using R23.(R23=(R2*R3)/(R2+R3)=16 ohms). Since R2 and R3 are in parallel, their voltages are the same. V23=R23*I1=40V. With this we can find the current through R2 and R3. I2=V23/R2=2A. I3=V23/R3=0.5A. a) I1=2.5A, I2=2A, I3=0.5A b) V1=10V, V2=V3=40V. c) Pdev=Vs*IT=125W Pabs=R1*I1^2+R2*I2^2+R3*I3^2=25+80+20=125W Pdev=Pabs=125W

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