Two resistors in series, R1 and R2, have a total resistance of 1,600 ohms. The resistance of R2 is 560 ohms less than twice R1. Find the resistance of R1 and R2.
Let's assign variables to the resistances of R1 and R2.
Let R1 be the resistance of R1.
Let R2 be the resistance of R2.
According to the given information, the total resistance of R1 and R2 in series is 1,600 ohms, so we can write the equation:
R1 + R2 = 1,600
The problem also states that the resistance of R2 is 560 ohms less than twice the resistance of R1, so we can write the equation:
R2 = 2R1 - 560
Now we can solve this system of equations to find the values of R1 and R2.
Substitute the second equation into the first equation:
R1 + (2R1 - 560) = 1600
Combine like terms:
3R1 - 560 = 1600
Add 560 to both sides:
3R1 = 1600 + 560
3R1 = 2160
Divide both sides by 3:
R1 = 2160 / 3
R1 = 720
Now substitute the value of R1 back into the equation for R2:
R2 = 2R1 - 560
R2 = 2(720) - 560
R2 = 1,440 - 560
R2 = 880
Therefore, the resistance of R1 is 720 ohms and the resistance of R2 is 880 ohms.
To find the resistance of R1 and R2, we can set up two equations based on the given information.
Let's assume the resistance of R1 is 'x' ohms.
According to the problem, the resistance of R2 is 560 ohms less than twice R1, which means it can be expressed as 2x - 560 ohms.
The total resistance of R1 and R2 in series is given as 1,600 ohms.
So, we can write the equation as follows:
x + (2x - 560) = 1,600
Now, let's solve this equation to find the value of 'x'.
Eq1: R1 + R2 = 1600 Ohms.
Eq2: R2 = 2R1 - 560 Ohms.
In Eq1, replace R2 with 2R1-560
R1 + 2R1-560 = 1600
3R1 = 2160
R1 = 720 Ohms
In Eq1, replace R1 with 720 Ohms.
720 + R2 = 1600
R2 = 880 Ohms