Questions: Attempt 1: 5 attempts remaining. If two resistors with resistances R1 and R2 are connected in parallel, as in the figure, then the total resistance R, measured in Ohms (Ω), is given by: 1/R = 1/R1 + 1/R2 If R1 and R2 are increasing at rates of .3 Ω / s and .2 Ω / s, respectively, how fast is R increasing when R1=80 Ω and R2=100 Ω ? Ω / s

Attempt 1: 5 attempts remaining.

If two resistors with resistances R1 and R2 are connected in parallel, as in the figure, then the total resistance R, measured in Ohms (Ω), is given by:
1/R = 1/R1 + 1/R2

If R1 and R2 are increasing at rates of .3 Ω / s and .2 Ω / s, respectively, how fast is R increasing when R1=80 Ω and R2=100 Ω ?
Ω / s

Transcript text: Attempt 1: 5 attempts remaining. If two resistors with resistances $R_{1}$ and $R_{2}$ are connected in parallel, as in the figure, then the total resistance $R$, measured in Ohms $(\Omega)$, is given by: \[ \frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}} \] If $R_{1}$ and $R_{2}$ are increasing at rates of $.3 \Omega / s$ and $.2 \Omega / s$, respectively, how fast is $R$ increasing when $R_{1}=80 \Omega$ and $R_{2}=100 \Omega$ ? $\Omega / s$ Submit answer Next item

Solution

Solution Steps

Step 1: Understand the Problem and Given Information

We are given two resistors in parallel with resistances $ R_1 $ and $ R_2 $. The formula for the total resistance $ R $ is: \[ \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} \] We need to find the rate of change of $ R $ with respect to time, $ \frac{dR}{dt} $, when $ R_1 = 80 \, \Omega $ and $ R_2 = 100 \, \Omega $, given that $ \frac{dR_1}{dt} = 0.3 \, \Omega/s $ and $ \frac{dR_2}{dt} = 0.2 \, \Omega/s $.

Step 2: Differentiate the Resistance Formula

Differentiate both sides of the equation with respect to time $ t $: \[ \frac{d}{dt}\left(\frac{1}{R}\right) = \frac{d}{dt}\left(\frac{1}{R_1} + \frac{1}{R_2}\right) \] Using the chain rule, we have: \[ -\frac{1}{R^2} \frac{dR}{dt} = -\frac{1}{R_1^2} \frac{dR_1}{dt} - \frac{1}{R_2^2} \frac{dR_2}{dt} \]

Step 3: Solve for $\frac{dR}{dt}$

Rearrange the equation to solve for $\frac{dR}{dt}$: \[ \frac{dR}{dt} = R^2 \left( \frac{1}{R_1^2} \frac{dR_1}{dt} + \frac{1}{R_2^2} \frac{dR_2}{dt} \right) \]

Step 4: Calculate $ R $ at Given Values

First, calculate $ R $ when $ R_1 = 80 \, \Omega $ and $ R_2 = 100 \, \Omega $: \[ \frac{1}{R} = \frac{1}{80} + \frac{1}{100} = \frac{5}{400} + \frac{4}{400} = \frac{9}{400} \] Thus, $ R = \frac{400}{9} \approx 44.4444 \, \Omega $.

Step 5: Substitute Values and Calculate $\frac{dR}{dt}$

Substitute the known values into the equation for $\frac{dR}{dt}$: \[ \frac{dR}{dt} = \left(\frac{400}{9}\right)^2 \left( \frac{1}{80^2} \times 0.3 + \frac{1}{100^2} \times 0.2 \right) \] Calculate each term: \[ \frac{1}{80^2} = \frac{1}{6400}, \quad \frac{1}{100^2} = \frac{1}{10000} \] \[ \frac{dR}{dt} = \left(\frac{160000}{81}\right) \left( \frac{0.3}{6400} + \frac{0.2}{10000} \right) \] \[ = \left(\frac{160000}{81}\right) \left( \frac{3}{64000} + \frac{2}{100000} \right) \] \[ = \left(\frac{160000}{81}\right) \left( \frac{15}{320000} + \frac{8}{400000} \right) \] \[ = \left(\frac{160000}{81}\right) \left( \frac{15 \times 5 + 8 \times 8}{1600000} \right) \] \[ = \left(\frac{160000}{81}\right) \left( \frac{75 + 64}{1600000} \right) \] \[ = \left(\frac{160000}{81}\right) \left( \frac{139}{1600000} \right) \] \[ = \frac{160000 \times 139}{81 \times 1600000} \] \[ = \frac{139}{81 \times 10} \] \[ = \frac{139}{810} \] \[ \approx 0.1716 \, \Omega/s \]