Respuesta :
Before starting with the problem, let's remind that:
- Total resistance of a combination of resistors in series is:
[tex]R_T=R_1+R_2+....[/tex]
- Total resistance of a combination of resistors in parallel is given by:
[tex]\frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}+...[/tex]
Now let's apply these equations to solve the different parts of the problem:
1. All the resistors in series: [tex]15.8 \Omega[/tex]
In this case, the total resistance is
[tex]R_T=6.50\Omega +7.60\Omega+1.70\Omega=15.8 \Omega[/tex]
2. All the resistors are connected in parallel: [tex]1.14 \Omega[/tex]
In this case, the total resistance is
[tex]\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[/tex]
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: [tex]5.20 \Omega[/tex]
The total resistance of the two resistors connected in parallel is
[tex]\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[/tex]
And the combination of these with the other resistor of 1.70-Ω in series gives a total resistance of
[tex]R_T=3.50 \Omega+1.70 \Omega=5.20 \Omega[/tex]
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: [tex]8.95 \Omega[/tex]
The total resistance of the two resistors connected in parallel is
[tex]\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[/tex]
And the combination of these with the other resistor of 7.60-Ω in series gives a total resistance of
[tex]R_T=1.35 \Omega+7.60 \Omega=8.95 \Omega[/tex]
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: [tex]7.89\Omega[/tex]
The total resistance of the two resistors connected in parallel is
[tex]\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[/tex]
And the combination of these with the other resistor of 6.50-Ω in series gives a total resistance of
[tex]R_T=1.39 \Omega+6.50 \Omega=7.89 \Omega[/tex]
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: [tex]1.52 \Omega[/tex]
In this case, the total resistance of the two resistors in series is
[tex]R_T=6.50\Omega + 7.60\Omega=14.10 \Omega[/tex]
And the combination of these with the other resistor of 1.70-Ω in parallel gives a total resistance of
[tex]\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[/tex]
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: [tex]3.94 \Omega[/tex]
In this case, the total resistance of the two resistors in series is
[tex]R_T=6.50\Omega + 1.70\Omega=8.20 \Omega[/tex]
And the combination of these with the other resistor of 7.60-Ω in parallel gives a total resistance of
[tex]\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[/tex]
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: [tex]3.83\Omega[/tex]
In this case, the total resistance of the two resistors in series is
[tex]R_T=7.60\Omega + 1.70\Omega=9.30 \Omega[/tex]
And the combination of these with the other resistor of 6.50-Ω in parallel gives a total resistance of
[tex]\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[/tex]
