Answer:
The total resistance of these three resistors connected in parallel is ![1.7143\Omega](https://tex.z-dn.net/?f=1.7143%5COmega)
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:
,
and
.
<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 (
,
and
).
![I_{R_1}=\frac{V_S}{R_1}](https://tex.z-dn.net/?f=I_%7BR_1%7D%3D%5Cfrac%7BV_S%7D%7BR_1%7D)
![I_{R_2}=\frac{V_S}{R_2}](https://tex.z-dn.net/?f=I_%7BR_2%7D%3D%5Cfrac%7BV_S%7D%7BR_2%7D)
![I_{R_3}=\frac{V_S}{R_3}](https://tex.z-dn.net/?f=I_%7BR_3%7D%3D%5Cfrac%7BV_S%7D%7BR_3%7D)
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}](https://tex.z-dn.net/?f=I_S%3DI_%7BR_1%7D%2BI_%7BR_2%7D%2BI_%7BR_3%7D%3D%5Cfrac%7BV_S%7D%7BR_1%7D%2B%5Cfrac%7BV_S%7D%7BR_2%7D%2B%5Cfrac%7BV_S%7D%7BR_3%7D%3D%5Cfrac%7BR_2%5Ccdot%20R_3%20%5Ccdot%20V_S%2BR_1%5Ccdot%20R_3%20%5Ccdot%20V_S%2BR_1%5Ccdot%20R_2%20%5Ccdot%20V_S%7D%7BR_1%5Ccdot%20R_2%5Ccdot%20R_3%7D)
![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}](https://tex.z-dn.net/?f=I_S%3DV_S%5Ccdot%20%5Cfrac%7BR_2%5Ccdot%20R_3%2BR_1%5Ccdot%20R_3%2BR_1%5Ccdot%20R_2%7D%7BR_1%5Ccdot%20R_2%5Ccdot%20R_3%7D)
The total resistance (
) is the source voltage divided by the source current:
![R_T=\frac{V_S}{I_S}](https://tex.z-dn.net/?f=R_T%3D%5Cfrac%7BV_S%7D%7BI_S%7D)
Now, replace
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}}](https://tex.z-dn.net/?f=R_T%3D%5Cfrac%7BV_S%7D%7BV_S%5Ccdot%20%5Cfrac%7BR_2%5Ccdot%20R_3%2BR_1%5Ccdot%20R_3%2BR_1%5Ccdot%20R_2%7D%7BR_1%5Ccdot%20R_2%5Ccdot%20R_3%7D%7D)
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}](https://tex.z-dn.net/?f=R_T%3D%5Cfrac%7BR_1%5Ccdot%20R_2%5Ccdot%20R_3%7D%7BR_2%5Ccdot%20R_3%2BR_1%5Ccdot%20R_3%2BR_1%5Ccdot%20R_2%7D)
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](https://tex.z-dn.net/?f=R_T%3D%5Cfrac%7B%286%5COmega%20%29%5Ccdot%20%283%5COmega%29%5Ccdot%20%2812%5COmega%29%7D%7B%283%5COmega%29%5Ccdot%20%2812%5COmega%29%2B%286%5COmega%29%5Ccdot%20%2812%5COmega%29%2B%286%5COmega%29%5Ccdot%20%283%5COmega%29%7D%3D%5Cfrac%7B12%7D%7B7%7D%5COmega%3D1.7143%5COmega%20)
Thus, the total resistance of these three resistors connected in parallel is ![1.7143\Omega](https://tex.z-dn.net/?f=1.7143%5COmega)
We have that x = 3, LM = 14 and LN = 2. options A, C and E
<h3>How to determine the value</h3>
We known that the three sides are on a straight line
LM + MN = LN
Let's substitute the values, we have
3x + 5 + 4x = 11x - 7
Collect like terms
3x + 4x - 11x = -7 -5
-4x = -12
x = -12/ -4
x = 3
LM = 3x + 5 = 3(3) + 5 = 14
LN = 11x - 7 = 11(3) -7 = 33 -7 = 26
Thus, we have that x = 3, LM = 14 and LN = 2. options A, C and E
Learn more about geometry here:
brainly.com/question/20303542
#SPJ1
Okay, the answer for the first problem is -24x-54. The second is -12k^4-16k^3+20k^2. Finally. the third one is -24x^3. Hope you get it.
Answer:
15/2
Step-by-step explanation:
The equation would be 425 x .83(12) + 250 because 425 is what he has at first, then times .83 which is what returns back to him each month, which .83 is multiplied by 12 which is how many months that are in a year plus 250 which is what he adds on after each year