Two resistors when connected in series to a 120-V line use one-fourth the power that is used when they are connected in parallel. If one resistor is 4.8 kohms. what is the resistance of the other?
E = 120 V.
R1 = 4.8k
R2 = ?
R1+R2 = 4(R1*R2)/(R1+R2)
(R1+R2)^2 = 4(R1*R2)
R1^2+2R1*R2+R2^2 = 4R1*R2
23.04+9.6R2+R2^2 = 19.2R2
R2^2 - 9.6R2 + 23.04 = 0
Use Quadratic formula
R2 = 4.8k
Check:
Parallel Connection
P1 = E^2/2.4k = 120^2/2.4k = 6,000 mW = 6 W.
Series connection
P2 = 120^2/(4.8k+4.8k) = 1,500 mW = 1.5 W.
P2/P1 = 1.5/6 = 1/4
Your answer is 4.8 k omes .
If one resistor is 4.8, then so is the other.
To solve this problem, let's denote the resistance of the second resistor as R.
We know that the power used in a circuit is given by the formula: P = V^2 / R, where P is power, V is voltage, and R is resistance.
When the resistors are connected in series, the power used is one-fourth of that used when they are connected in parallel. Therefore, we can write the following equation:
P(series) = 1/4 * P(parallel)
Using the formulas for power, we can substitute in the given values:
V^2 / (4.8 kohms + R) = 1/4 * V^2 / (4.8 kohms * R)
Simplifying the equation:
(4.8 kohms + R) = 1/4 * (4.8 kohms * R)
(4.8 kohms + R) = (4.8 kohms * R) / 4
Now, let's solve for R:
4.8 kohms + R = (4.8 kohms * R) / 4
Multiplying both sides of the equation by 4:
19.2 kohms + 4R = 4.8 kohms * R
Rearranging the equation:
4R - 4.8 kohms * R = -19.2 kohms
Factoring out R:
R * (4 - 4.8 kohms) = -19.2 kohms
Dividing both sides of the equation by (4 - 4.8 kohms):
R = -19.2 kohms / (4 - 4.8 kohms)
Since resistance cannot be negative, we can discard the negative sign and take the absolute value:
R = 19.2 kohms / (4 - 4.8 kohms)
Calculating the value:
R = 19.2 kohms / (4 - 4.8 kohms)
R = 19.2 kohms / (-0.8 kohms)
R = -24 kohms
Therefore, the resistance of the other resistor is 24 kohms.
To solve this problem, let's first understand the concept of power in resistors.
The power (P) in a resistor can be calculated using the formula:
P = (V^2) / R
where P is power, V is voltage, and R is resistance.
When the resistors are connected in series, the total resistance (Rs) can be found by adding the individual resistances:
Rs = R1 + R2
The power consumed when the resistors are in series is given as one-fourth of the power consumed when they are in parallel:
Pseries = (1/4) * Pparallel
Now, let's calculate the values step-by-step:
1. Given that one resistor is 4.8 kilohms (kΩ), we can assume R1 = 4.8 kΩ.
2. Let's find the voltage (V) in both cases. Since the resistors are connected to a 120-V line, V is the same in both cases. Therefore, Vseries = Vparallel = 120 V.
3. Now, we can calculate the power (Pseries) consumed by the resistors when connected in series:
Pseries = (Vseries^2) / Rs
4. We can calculate the power (Pparallel) consumed by the resistors when connected in parallel:
Pparallel = (Vparallel^2) / (R1 + R2)
5. Given that Pseries = (1/4) * Pparallel, we can set up the equation:
(Vseries^2) / Rs = (1/4) * ((Vparallel^2) / (R1 + R2))
6. Substitute the known values into the equation:
(120^2) / Rs = (1/4) * ((120^2) / (4.8 kΩ + R2))
7. Simplify the equation:
Rs = (1/4) * (120^2) / (120^2) / (4.8 kΩ + R2)
8. Cross-multiply the equation:
(120^2) * (4.8 kΩ + R2) = (1/4) * (120^2) * Rs
9. Cancel out the common terms:
4.8 kΩ + R2 = (1/4) * Rs
10. Multiply through by 4 to get rid of the fraction:
19.2 kΩ + 4R2 = Rs
11. Now, substitute the values and solve for R2:
19.2 kΩ + 4R2 = R1 + R2
19.2 kΩ + 4R2 = 4.8 kΩ + R2
3R2 = 14.4 kΩ
R2 = 14.4 kΩ / 3
R2 = 4.8 kΩ
Therefore, the resistance of the other resistor, when one resistor is 4.8 kilohms, is also 4.8 kilohms.