Solve by setting up the proper equation to describe the facts given and then carrying out the mathematical calculations to solve for the unknown variable(s).
Three batteries are connected in series so that the total voltage is 54 volts. The voltage of the first battery is twice the voltage of the second and 1/3 the voltage of the third battery. Find the actual voltage of each battery.
Show answer in the following format #,#,#.
Va+Vb+Vc=54
The voltage of the first battery is twice the voltage of the second and 1/3 the voltage of the third battery. This statement is not definitive.
does it mean Va=2Vb AND Va=1/3 Vc or does it mean
Va=2Va+1/3 Vc OR does it mean
Va=2(Va+1/3 Vc).
The problem cannot be solved until that is known.
nonsense. let the three batteries by A,B,C
then A = 2B
3A = C
A + B + C = 54
so A = 12
B = 6
C = 36
Let's define the unknown voltage of the second battery as x volts.
Based on the given information, the voltage of the first battery is twice the voltage of the second battery, so it is 2x volts. The voltage of the third battery is 1/3 the voltage of the first battery, so it is (1/3) * (2x) = (2/3)x volts.
Since the three batteries are connected in series, the total voltage is the sum of the individual voltages:
2x + x + (2/3)x = 54
Combining like terms:
5x + (2/3)x = 54
Multiplying through by 3 to eliminate the fraction:
15x + 2x = 162
17x = 162
Dividing both sides by 17 to solve for x:
x = 162 / 17
x ≈ 9.529
Now we can find the actual voltage of each battery:
First battery: 2x = 2 * 9.529 ≈ 19.06 volts
Second battery: x ≈ 9.529 volts
Third battery: (2/3)x = (2/3) * 9.529 ≈ 6.352 volts
So, the actual voltage of each battery is approximately 19.06 volts, 9.529 volts, and 6.352 volts respectively.
Answer: 19.06, 9.529, 6.352.
To solve this problem, let's assume the voltage of the second battery is "x" volts.
According to the given information, the voltage of the first battery is twice the voltage of the second battery, so the voltage of the first battery is 2x volts.
Similarly, the voltage of the third battery is 1/3 times the voltage of the first battery, so the voltage of the third battery is (1/3)(2x) = (2/3)x volts.
Now, we can set up an equation to represent the total voltage in the series connection:
2x + x + (2/3)x = 54
Combining like terms:
(6/3)x + (3/3)x + (2/3)x = 54
(11/3)x = 54
To solve for x, we'll multiply both sides of the equation by the reciprocal of (11/3), which is (3/11):
x = (54)(3/11)
x ≈ 14.7273
So, the voltage of the second battery is approximately 14.7273 volts.
Now, we can find the voltage of the first battery:
2x = 2(14.7273) ≈ 29.4546 volts
Finally, we can find the voltage of the third battery:
(2/3)x = (2/3)(14.7273) ≈ 9.8182 volts
Therefore, the actual voltage of each battery is approximately 29.4546 volts, 14.7273 volts, and 9.8182 volts.
The answer in the requested format is: 29.4546, 14.7273, 9.8182.