Solved: Resistors connected in parallel If two resistors

Chapter , Problem 145PE

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QUESTION:

Resistors connected in parallel If two resistors of \(R_{1}\) and \(R_{2}\) ohms are connected in parallel in an electric circuit to make an \(R\) - ohm resistor, the value of \(R\) can be found from the equation

\(\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}\).

If \(R_{1}\) is decreasing at the rate of 1 ohm sec and \(R_{2}\) is increasing at the rate of 0.5 ohm sec, at what rate is \(R\) changing when \(R_{1}=75\) ohms and \(R_{2}=50\)ohms?

Equation Transcription:

 

Text Transcription:

R_1

R_2

R

R

frac{1}{R} = frac{1}{R_1} + frac{1}{R_2}

R_1

R_2

R

R_1 = 75

R_2 = 50

Questions & Answers

QUESTION:

Resistors connected in parallel If two resistors of \(R_{1}\) and \(R_{2}\) ohms are connected in parallel in an electric circuit to make an \(R\) - ohm resistor, the value of \(R\) can be found from the equation

\(\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}\).

If \(R_{1}\) is decreasing at the rate of 1 ohm sec and \(R_{2}\) is increasing at the rate of 0.5 ohm sec, at what rate is \(R\) changing when \(R_{1}=75\) ohms and \(R_{2}=50\)ohms?

Equation Transcription:

 

Text Transcription:

R_1

R_2

R

R

frac{1}{R} = frac{1}{R_1} + frac{1}{R_2}

R_1

R_2

R

R_1 = 75

R_2 = 50

ANSWER:

Solution:-

Step 1 of 4

Given that

 We have to find the related rates of the connected resistance if R1 is decreasing at the rate of 1 ohm sec and R2 is increasing at the rate of 0.5 ohm sec, when R1 = 75 ohms and R2 = 50 ohms

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