Physics, 05.12.2019 12:31, maribel234

Eight different values of resistance can be obtained by connecting together three resistors 6.50 ω, 7.60 ω, and 1.70 ω in all possible ways. what are the values in the following situations? all the resistors are connected in series. ω all the resistors are connected in parallel. ω the 6.50 ω and 7.60 ω resistors are connected in parallel, and the 1.70-ω resistor is connected in series with the parallel combination. ω the 6.50 ω and 1.70 ω resistors are connected in parallel, and the 7.60-ω resistor is connected in series with the parallel combination. ω the 7.60 ω and 1.70 ω resistors are connected in parallel, and the 6.50-ω resistor is connected in series with the parallel combination. ω the 6.50 ω and 7.60 ω resistors are connected in series, and the 1.70-ω resistor is connected in parallel with the series combination. ω the 6.50 ω and 1.70 ω resistors are connected in series, and the 7.60-ω resistor is connected in parallel with the series combination. ω the 7.60 ω and 1.70 ω resistors are connected in series, and the 6.50-ω resistor is connected in parallel with the series combination. ω additional materials

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answered: Amyra2003

Before starting with the problem, let's remind that:

- Total resistance of a combination of resistors in series is:

R_T=R_1+R_2+....

- Total resistance of a combination of resistors in parallel is given by:

$\frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}+...$

Now let's apply these equations to solve the different parts of the problem:

1. All the resistors in series: $15.8 \Omega$

In this case, the total resistance is

$R_T=6.50\Omega +7.60\Omega+1.70\Omega=15.8 \Omega$

2. All the resistors are connected in parallel: $1.14 \Omega$

In this case, the total resistance is

$\frac{1}{R_T}=\frac{1}{6.50 \Omega}+\frac{1}{7.60 \Omega}+\frac{1}{1.70 \Omega}=0.874\Omega^{-1}\\R_T=\frac{1}{0.874 \Omega^{-1}}=1.14 \Omega$

3. The 6.50 Ω and 7.60 Ω resistors are connected in parallel, and the 1.70-Ω resistor is connected in series with the parallel combination: $5.20 \Omega$

The total resistance of the two resistors connected in parallel is

$\frac{1}{R_T}=\frac{1}{6.50 \Omega}+\frac{1}{7.60\Omega}=0.285\Omega^{-1}\\R_T=\frac{1}{0.285 \Omega^{-1}}=3.50 \Omega$

And the combination of these with the other resistor of 1.70-Ω in series gives a total resistance of

$R_T=3.50 \Omega+1.70 \Omega=5.20 \Omega$

4. The 6.50 Ω and 1.70 Ω resistors are connected in parallel, and the 7.60-Ω resistor is connected in series with the parallel combination: $8.95 \Omega$

The total resistance of the two resistors connected in parallel is

$\frac{1}{R_T}=\frac{1}{6.50 \Omega}+\frac{1}{1.70\Omega}=0.742\Omega^{-1}\\R_T=\frac{1}{0.742 \Omega^{-1}}=1.35 \Omega$

And the combination of these with the other resistor of 7.60-Ω in series gives a total resistance of

$R_T=1.35 \Omega+7.60 \Omega=8.95 \Omega$

5. The 7.60 Ω and 1.70 Ω resistors are connected in parallel, and the 6.50-Ω resistor is connected in series with the parallel combination: $7.89\Omega$

The total resistance of the two resistors connected in parallel is

$\frac{1}{R_T}=\frac{1}{7.60 \Omega}+\frac{1}{1.70\Omega}=0.720\Omega^{-1}\\R_T=\frac{1}{0.720 \Omega^{-1}}=1.39 \Omega$

And the combination of these with the other resistor of 6.50-Ω in series gives a total resistance of

$R_T=1.39 \Omega+6.50 \Omega=7.89 \Omega$

6. The 6.50 Ω and 7.60 Ω resistors are connected in series, and the 1.70-Ω resistor is connected in parallel with the series combination: $1.52 \Omega$

In this case, the total resistance of the two resistors in series is

$R_T=6.50\Omega + 7.60\Omega=14.10 \Omega$

And the combination of these with the other resistor of 1.70-Ω in parallel gives a total resistance of

$\frac{1}{R_T}=\frac{1}{14.10 \Omega}+\frac{1}{1.70\Omega}=0.659\Omega^{-1}\\R_T=\frac{1}{0.720 \Omega^{-1}}=1.52 \Omega$

7. The 6.50 Ω and 1.70 Ω resistors are connected in series, and the 7.60-Ω resistor is connected in parallel with the series combination: $3.94 \Omega$

In this case, the total resistance of the two resistors in series is

$R_T=6.50\Omega + 1.70\Omega=8.20 \Omega$

And the combination of these with the other resistor of 7.60-Ω in parallel gives a total resistance of

$\frac{1}{R_T}=\frac{1}{8.20 \Omega}+\frac{1}{7.6\Omega}=0.254\Omega^{-1}\\R_T=\frac{1}{0.720 \Omega^{-1}}=3.94 \Omega$

8. The 7.60 Ω and 1.70 Ω resistors are connected in series, and the 6.50-Ω resistor is connected in parallel with the series combination: $3.83\Omega$

In this case, the total resistance of the two resistors in series is

$R_T=7.60\Omega + 1.70\Omega=9.30 \Omega$

And the combination of these with the other resistor of 6.50-Ω in parallel gives a total resistance of

$\frac{1}{R_T}=\frac{1}{9.30 \Omega}+\frac{1}{6.50\Omega}=0.261\Omega^{-1}\\R_T=\frac{1}{0.720 \Omega^{-1}}=3.83 \Omega$

answered: Guest

Answer: it will take 4.36 seconds to reach her average velocity 10.69 miles/hour.explanation: we have equation of motion $s=ut+\frac{1}{2} at^2$ ,where s is the displacement, a is the acceleration, u is the initial velocity and t is the time taken.average velocity = displacement/ time taken.average velocity = 10.69 miles/hr = 10.69*1.6*1000/(60*60) = 4.75 m/s4.75 = s/ts = 4. 1we also have acceleration = 2.18 m/s², initial velocity = 0 m/s $s=0*t+\frac{1}{2} *2.18*t^2$ $s=1.09t^2[/ 2equating equation 1 and 2 [tex]4.75t = 1.09t^2\\ \\ t=4.36 seconds.$ so it will take 4.36 seconds to reach her average velocity 10.69 miles/hour.

answered: Guest

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