When two resistors are connected in series across a 12-V battery, 1.1A flows. When they are connected in parallel across the same battery, 6.3A flows. What is the smaller resistor's value?
V/rparallel=6.35
V/rsum=1.2
rsum=12/1.2=you do it.
R1+R2=rsum=above
then r2=abovesum-r1
rparallel=12/6.35
or
r1r2/(r1+r2)=12/6.35
r1(abovesum-r1)=12/6.35 (r1+12-r1)
and that quickly solves as a quadratic.
check my math, I did it in my head.
To find the value of the smaller resistor, we will use the formulas for resistors in series and in parallel.
Let's assume that the resistance of the larger resistor is R1 and the resistance of the smaller resistor is R2.
When two resistors are connected in series, the total resistance (RT) is given by:
RT = R1 + R2
When two resistors are connected in parallel, the total resistance (RT) is given by:
1/RT = 1/R1 + 1/R2
Given that the current flowing in the series circuit (IS) is 1.1A and the current flowing in the parallel circuit (IP) is 6.3A, and the battery voltage (V) is 12V, we can use Ohm's Law (V = I * R) to derive the resistance values:
In the series circuit:
V = IS * RT
12V = 1.1A * (R1 + R2) ----(1)
In the parallel circuit:
V = IP * RT
12V = 6.3A * 1/RT
12V = 6.3A * 1/(1/R1 + 1/R2) ----(2)
Simplifying equation (2):
12V = 6.3A * R1 * R2 / (R1 + R2)
12V * (R1 + R2) = 6.3A * R1 * R2
12R1 + 12R2 = 6.3R1R2 ----(3)
Now, we have two equations (equations 1 and 3) with two unknowns (R1 and R2). We can solve these equations simultaneously to find the values of R1 and R2.
Using equation (1):
12V = 1.1A * (R1 + R2)
12V = 1.1A * R1 + 1.1A * R2
12V - 1.1A * R1 = 1.1A * R2
R2 = (12V - 1.1A * R1) / 1.1A ----(4)
Substituting equation (4) into equation (3):
12R1 + 12((12V - 1.1A * R1) / 1.1A) = 6.3R1((12V - 1.1A * R1) / 1.1A)
12R1 + (12 * (12V - 1.1A * R1)) / 1.1A = 6.3R1((12V - 1.1A * R1) / 1.1A)
Simplifying and rearranging the equation:
12R1 + (144V - 15.84A * R1) / 1.1A = 6.3R1((12V - 1.1A * R1) / 1.1A)
12R1 + (1440V - 158.4A * R1) = 6.3R1(12V - 1.1A * R1)
Expanding and simplifying further:
12R1 + 1440V - 158.4A * R1 = 72.6V * R1 - 7.29A * R1^2
This equation is a quadratic equation. We can solve it to find the values of R1, and then substitute R1 back into equation (4) to find R2.
To find the value of the smaller resistor, we can use Ohm's law, which states that the current flowing through a resistor is equal to the voltage across it divided by its resistance.
Let's call the resistance of the smaller resistor R1 and the resistance of the larger resistor R2.
Using the information given:
- When the two resistors are connected in series, the total current flowing is 1.1A.
- When the two resistors are connected in parallel, the total current flowing is 6.3A.
- The voltage across the resistors is the same in both cases and is equal to 12V.
Now we can set up equations using Ohm's law to solve for the resistances:
For the series connection:
V = I1 * R1 + I2 * R2
The total current (I) is 1.1A, and the voltage (V) is 12V. Since the resistors are in series, the current is the same through both resistors (I1 = I2 = 1.1A). Substituting these values into the equation:
12V = 1.1A * R1 + 1.1A * R2
12V = 1.1A * (R1 + R2) ---(Equation 1)
For the parallel connection:
V = I1 * R1
The total current (I) is 6.3A, and the voltage (V) is 12V. Substituting these values into the equation:
12V = 6.3A * R1
R1 = 12V / 6.3A
R1 = 1.90Ω ---(Equation 2)
We now have two equations involving the resistances R1 and R2. We can solve these equations simultaneously to find the value of R2.
Substituting Equation 2 into Equation 1:
12V = 1.1A * (1.90Ω + R2)
12V = 2.09Ω + 1.1A * R2
1.1A * R2 = 12V - 2.09Ω
R2 = (12V - 2.09Ω) / 1.1A
R2 ≈ 8.37Ω
Therefore, the smaller resistor's value is approximately 1.90Ω.