What is the x-value of the solution to this system of equations?

1 answer:

4 0

Plug in for x.

7(2y - 4) + 5y = -66

14y - 28 + 5y = -66

19y - 28 = -66

19y = -38

y = -2

Put it back into the equation.
2(-2) - 4 = -8
x = -8

The x-value is x = -8

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What is the total resistance of a parallel circuit that has three loads? Load one has a resistance of 6 ohms. Load two has a res

gizmo_the_mogwai [7]

Answer:

The total resistance of these three resistors connected in parallel is $1.7143\Omega$

Step-by-step explanation:

The attached image has the circuit for finding the total resistance. The circuit is composed by a voltage source and three resistors connected in parallel: $R_1=6\Omega$ , $R_2=3\Omega$ and $R_3=12\Omega$ .

<u>First step: to find the source current</u>

The current that the source provides is the sum of the current that each resistor consumes. Keep in mind that the voltage is the same for the three resistors ( $R_1$ , $R_2$ and $R_3$ ).

$I_{R_1}=\frac{V_S}{R_1}$

$I_{R_2}=\frac{V_S}{R_2}$

$I_{R_3}=\frac{V_S}{R_3}$

The total current is:

$I_S=I_{R_1}+I_{R_2}+I_{R_3}=\frac{V_S}{R_1}+\frac{V_S}{R_2}+\frac{V_S}{R_3}=\frac{R_2\cdot R_3 \cdot V_S+R_1\cdot R_3 \cdot V_S+R_1\cdot R_2 \cdot V_S}{R_1\cdot R_2\cdot R_3}$

$I_S=V_S\cdot \frac{R_2\cdot R_3+R_1\cdot R_3+R_1\cdot R_2}{R_1\cdot R_2\cdot R_3}$

The total resistance ( $R_T$ ) is the source voltage divided by the source current:

$R_T=\frac{V_S}{I_S}$

Now, replace $I_S$ by the previous expression and the total resistance would be:

$R_T=\frac{V_S}{V_S\cdot \frac{R_2\cdot R_3+R_1\cdot R_3+R_1\cdot R_2}{R_1\cdot R_2\cdot R_3}}$

Simplify the expression and you must get:

$R_T=\frac{R_1\cdot R_2\cdot R_3}{R_2\cdot R_3+R_1\cdot R_3+R_1\cdot R_2}$

The last step is to replace the values of the resistors:

$R_T=\frac{(6\Omega )\cdot (3\Omega)\cdot (12\Omega)}{(3\Omega)\cdot (12\Omega)+(6\Omega)\cdot (12\Omega)+(6\Omega)\cdot (3\Omega)}=\frac{12}{7}\Omega=1.7143\Omega$