Answer :
Two resistors are connected in parallel. If R1 and R2 represent the resistance in Ohms (Ω) of each resistor, then the total resistance R is given by [tex]\mathbf{\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}}[/tex]. Thus, the rate of R changes when R₁ = 117 Ω and
R₂ = 112 Ω is 0.25 Ω/min
For a given resistor connected in parallel;
[tex]\mathbf{\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}}[/tex]
Making R from the left-hand side the subject of the formula, then:
[tex]\mathbf{R = \dfrac{R_1R_2}{R_1+R_2}}[/tex]
Given that:
- [tex]\mathbf{R_1 = 117,}[/tex]
- [tex]\mathbf{R_2 = 112 }[/tex]
Now, replacing the values in the above previous equation, we have:
[tex]\mathbf{R = \dfrac{13104}{229}}[/tex]
However, the differentiation of R with respect to time t will give us the rate at which R is changing when R1=117Ω and R2=112Ω.
So, by differentiating the given equation of the resistor in parallel with respect to time t;
[tex]\mathbf{\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}}[/tex], we have:
[tex]\mathbf{\dfrac{1}{R^2}(\dfrac{dR}{dt})=\dfrac{1}{R_1^2}(\dfrac{dR_1}{dt})+\dfrac{1}{R_2^2}(\dfrac{dR_2}{dt})}[/tex]
[tex]\mathbf{(\dfrac{dR}{dt})=R^2 \Bigg[ \dfrac{1}{R_1^2}(\dfrac{dR_1}{dt})+\dfrac{1}{R_2^2}(\dfrac{dR_2}{dt})\Bigg]}[/tex]
[tex]\mathbf{\dfrac{dR}{dt}=(\dfrac{13104}{229})^2 \Bigg[ \dfrac{0.4}{117^2}+\dfrac{0.6}{112^2}\Bigg]}[/tex]
[tex]\mathbf{\dfrac{dR}{dt}=3274.44 \Bigg[ (7.7052 \times 10^{-5} )\Bigg]}[/tex]
[tex]\mathbf{\dfrac{dR}{dt}=0.25\ \Omega /min}[/tex]
Therefore, we can conclude that the rate at which R is changing R1=117Ω and R2=112Ω is 0.25 Ω/min
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