Two resistances, R1 and R2, are connected in series across a 15-V battery. The current increases by 0.050 A when R2 is removed, leaving R1 connected across the battery. However, the current increases by just 0.025 A when R1 is removed, leaving R2 connected across the battery.
Find R1 and R2
To find the values of R1 and R2, we can use Ohm's Law and the relationship between current, voltage, and resistance.
Let's start by analyzing the situation when both R1 and R2 are connected in series across the 15V battery.
1. When both resistances are connected in series, the total current can be calculated using Ohm's Law:
I_total = V / (R1 + R2) ... (Equation 1)
2. Next, let's determine the change in current when we remove either R1 or R2.
a. When R2 is removed, the current becomes:
I_R1 = V / R1 ... (Equation 2)
b. When R1 is removed, the current becomes:
I_R2 = V / R2 ... (Equation 3)
According to the problem statement, the current increases by 0.050A when R2 is removed (leaving R1 connected), and the current increases by 0.025A when R1 is removed (leaving R2 connected).
Using this information, we can set up the following two equations:
Equation 2 - Equation 1: I_R1 - I_total = 0.050A
(V / R1) - (V / (R1 + R2)) = 0.050A
Equation 3 - Equation 1: I_R2 - I_total = 0.025A
(V / R2) - (V / (R1 + R2)) = 0.025A
Now, let's solve these equations to find the values of R1 and R2.
Simplifying Equation 2:
(V / R1) - (V / (R1 + R2)) = 0.050A
(V(R1 + R2) - VR1) / (R1(R1 + R2)) = 0.050A
VR2 / (R1(R1 + R2)) = 0.050A
Simplifying Equation 3:
(V / R2) - (V / (R1 + R2)) = 0.025A
(V(R1 + R2) - VR2) / (R2(R1 + R2)) = 0.025A
VR1 / (R2(R1 + R2)) = 0.025A
Now, let's set up a system of equations:
VR2 / (R1(R1 + R2)) = 0.050A ... (Equation 4)
VR1 / (R2(R1 + R2)) = 0.025A ... (Equation 5)
From Equation 4, we can isolate VR2:
VR2 = 0.050A * R1 * (R1 + R2)
From Equation 5, we can isolate VR1:
VR1 = 0.025A * R2 * (R1 + R2)
Now, we can substitute these values back into Equations 4 and 5:
0.050A * R1 * (R1 + R2) / (R1(R1 + R2)) = 0.050A
0.025A * R2 * (R1 + R2) / (R2(R1 + R2)) = 0.025A
Simplifying the equations:
0.050A * R1 * (R1 + R2) = 0.050A * R1 * (R1 + R2)
0.025A * R2 * (R1 + R2) = 0.025A * R2 * (R1 + R2)
Since both sides of the equations are equal, we can remove the denominators and continue solving:
0.050A * R1 * (R1 + R2) = 0.050A * R1 * (R1 + R2)
0.025A * R2 * (R1 + R2) = 0.025A * R2 * (R1 + R2)
Simplifying further:
0.050(R1 + R2) = 0.050(R1 + R2)
0.025(R1 + R2) = 0.025(R1 + R2)
Since R1 + R2 is common to both equations, we can conclude that R1 + R2 is equal in both cases.
Now, dividing Equation 4 by Equation 5 yields:
(0.050A * R1 * (R1 + R2))/(0.025A * R2 * (R1 + R2)) = 2
Canceling out the common terms, we get:
2R1 = R2
Given this relationship, we can express R2 in terms of R1:
R2 = 2R1
Substituting this into Equation 1:
I_total = V / (R1 + R2)
I_total = V / (R1 + 2R1)
I_total = V / 3R1
We also know from Equation 2 that:
I_R1 - I_total = 0.050A
Substituting the value of I_total:
(V / R1) - (V / 3R1) = 0.050A
Multiplying both sides by 3R1:
3V - V = 0.150A * R1
2V = 0.150A * R1
R1 = (2V) / (0.150A)
Finally, substituting this value of R1 into the expression for R2:
R2 = 2R1
R2 = 2 * [(2V) / (0.150A)]
Simplifying further yields the values of R1 and R2.
To find the values of R1 and R2, we can use a couple of key equations related to resistances and current in a series circuit.
First, let's use the fact that the current increases by 0.050 A when R2 is removed. We can call the current with R2 connected I2, and the current without R2 connected I1. We know that:
I2 - I1 = 0.050 A
Next, when R1 is removed, the current increases by 0.025 A. Again, let's call the current with R1 connected I1, and the current without R1 connected I2. We have:
I1 - I2 = 0.025 A
Now, let's use Ohm's Law to write the equations in terms of resistance. Ohm's Law states that V = IR, where V is the voltage across the resistor, I is the current flowing through the resistor, and R is the resistance of the resistor.
For R2, when it is connected, the voltage across R2 will be 15 V. So we have:
I1 * R2 = 15
When R2 is removed, R1 remains connected, so the voltage across R1 is still 15 V. This gives us:
I2 * R1 = 15
Now we have two equations and two unknowns (R1 and R2), and we can solve for their values.
From the first equation, we can rearrange it to solve for I2:
I2 = I1 - 0.050
Plugging this into the second equation:
(I1 - 0.050) * R1 = 15
Now we can substitute the first equation into the second equation:
(I1 - 0.050) * R1 = 15
I1 * R1 - 0.050 * R1 = 15
I1 * R1 = 15 + 0.050 * R1
Now we substitute the second equation into the first equation:
(I1 - 0.025) * R2 = 15
I1 * R2 - 0.025 * R2 = 15
I1 * R2 = 15 + 0.025 * R2
Now we have a system of equations to solve simultaneously:
(1) I1 * R1 = 15 + 0.050 * R1
(2) I1 * R2 = 15 + 0.025 * R2
Using the system of equations, we can rearrange for R1:
R1 = (15 + 0.050 * R1) / I1
And rearrange for R2:
R2 = (15 + 0.025 * R2) / I1
To find the values of R1 and R2, we need the values of I1 (current with R1) and I2 (current with R2). You can measure these currents using an ammeter or obtain them from the problem statement. Substituting these values into the equations will give you the values of R1 and R2.
V= IR
in this case V = I(R1+R2)
so we have 15 = IR1 + IR2
(15 - IR1)/I = R2 we will use this later in our subsitution.
We also know that...
15 =(I+.05)R1 and 15 = (I+.025)R2
since both equal 15, they must equal each other:
(I+.05)R1 = (I+.025)R2
Use the substitution for R2 from above and you will be able to solve for R1. by eliminating the I.