Two identical resistors are connected in parallel across a 26-V battery, which supplies them with a total power of 9.8 W. While the battery is still connected, one of the resistors is heated so that its resistance doubles. The resistance of the other resistor remains unchanged.
Answer:
V^2/watts = ohms
V^2/ohms + V^2/2(ohms) = W
power = V^2/Req = 9.8 = 26^2/Req
so
Req = 69 ohms
so R = 2 Req = 138 Ohms
new Req:
1/Req = 1/138 + 1/276
Req = 92 Ohms
power = 26^2/92 = 7.35 Watts
First, let's calculate the initial resistance of the two identical resistors connected in parallel. Since they are identical, their resistances are equal.
Let's denote the resistance of each resistor as R. In a parallel connection, the equivalent resistance (Re) can be calculated using the formula:
1/Re = 1/R + 1/R
Simplifying this equation, we get:
1/Re = 2/R
To find Re, we take the reciprocal of both sides:
Re = R/2
Now, let's calculate the initial power supplied by the battery. We are given that the total power is 9.8 W.
The power (P) in a circuit can be calculated using the formula:
P = V^2 / R
Substituting the given values, we get:
9.8 = 26^2 / Re
Now, let's substitute Re with R/2:
9.8 = 26^2 / (R/2)
Next, let's simplify this equation:
9.8 = (676)/(R/2)
To get rid of the fraction, we multiply both sides by (R/2):
9.8 * (R/2) = 676
4.9 * R = 676
Now, let's solve for R:
R = 676 / 4.9
R ≈ 138.37 ohms
Now that we have the initial resistance of each resistor, we can calculate the initial voltage across each resistor using Ohm's law:
V = √(P * R)
For one resistor, we will use the initial resistance (R ≈ 138.37 ohms) and power (P = 9.8 W):
V1 = √(9.8 * 138.37) ≈ 38.194 V
For the other resistor, we know that its resistance remains unchanged. Therefore, the voltage across this resistor will also be approximately 38.194 V.
Now, let's consider the scenario where one of the resistors is heated so that its resistance doubles. Let's call the new resistance of this resistor R'.
Since the other resistor's resistance remains unchanged, let's call it R as before.
Now, let's calculate the new values.
We are given that R' is double the resistance of R.
R' = 2R
Replacing R with 138.37 ohms:
R' = 2 * 138.37 ≈ 276.74 ohms
Now, let's calculate the new total resistance when one resistor has a resistance of R' and the other has a resistance of R.
The new equivalent resistance (Re') can be calculated using the formula:
1/Re' = 1/R' + 1/R
Substituting the values, we get:
1/Re' = 1/(276.74) + 1/(138.37)
Simplifying this equation, we get:
1/Re' ≈ 0.00361 + 0.00722
1/Re' ≈ 0.01083
To find Re', we take the reciprocal of both sides:
Re' ≈ 1/0.01083
Re' ≈ 92.31 ohms
Now, let's calculate the new power supplied by the battery. We are still given that the total power is 9.8 W.
Using the formula:
P = V^2 / Re'
Substituting the given values, we get:
9.8 = 26^2 / 92.31
Now, let's solve for V^2:
V^2 = 9.8 * 92.31
V^2 ≈ 904.298
Finally, let's calculate the new voltage (V') across each resistor using the value of V^2:
V' = √V^2
V' ≈ √904.298
V' ≈ 30.076 V
So, the new voltage across each resistor is approximately 30.076 V.
To find the value of the resistance of each resistor, we can use the power equation:
P = V^2 / R
where P is the power in watts, V is the voltage in volts, and R is the resistance in ohms.
Given that the total power supplied by the battery is 9.8 W and the voltage is 26 V, we can plug these values into the equation:
9.8 = 26^2 / (R1 + R2)
Since the resistors are identical, let's denote their resistance as R. Therefore, R1 = R and R2 = R.
9.8 = (26^2) / (R + R)
9.8 = 676 / (2R)
Multiplying both sides by (2R), we get:
(2R) * 9.8 = 676
19.6R = 676
R = 34.49 ohms (approximately)
So, each resistor has a resistance of 34.49 ohms when they are connected in parallel across the 26-V battery.
Now, one of the resistors is heated so that its resistance doubles. Let's call the new resistance of this resistor as R1'. The resistance of the other resistor remains unchanged, so the resistance of the second resistor is still R.
Therefore, the total resistance of the circuit, when one of the resistors has double the resistance, can be calculated as:
Total resistance (R_total) = R1' + R
Since R1' is double the initial resistance (2R), we have:
R_total = 2R + R
R_total = 3R
Now we can substitute the value of R from earlier:
R_total = 3 * 34.49 ohms
R_total = 103.47 ohms (approximately)
Thus, when one of the resistors has its resistance doubled, the total resistance of the circuit becomes approximately 103.47 ohms.