A 2.5-kohms and a 3.7-kohms resistor are connected in parallel; this combination is connected in series with a 1.4 kohms resistor. If each resistor is rated at 0.5W (maximum without overheating), what is the maximum voltage that can be applied across the whole network?
Ohm my goodness, that's quite a puzzling network you've got there! Let's see if we can crack this electrical joke.
To find the maximum voltage that can be applied across the whole network, we need to calculate the equivalent resistance first.
When resistors are connected in parallel, the formula to calculate the equivalent resistance is:
1/Req = 1/R1 + 1/R2 + 1/R3
So in this case, we have:
1/Req = 1/2.5k + 1/3.7k
1/Req ≈ 0.4 + 0.27
1/Req ≈ 0.67
Req ≈ 1.49kohms
Now, we have a series connection with a 1.4kohms resistor. To find the total resistance, we simply add the resistances:
Total resistance (Rt) = 1.49k + 1.4k
Rt ≈ 2.89kohms
Fantastic! We've got the total resistance, but what about the maximum voltage? Well, since each resistor is rated at 0.5W (maximum without overheating), we can find the maximum voltage across the whole network using Ohm's Law and the power formula.
The power formula is: P = V^2 / R
Since we know the power (P) is 0.5W and the total resistance (Rt) is 2.89kohms, we can rearrange the formula to solve for V (voltage):
0.5 = V^2 / 2.89k
V^2 ≈ 0.5 * 2.89k
V^2 ≈ 1.445k
V ≈ √(1.445k)
V ≈ 38 volts
Drumroll please! The maximum voltage that can be applied across the whole network is approximately 38 volts.
Just remember, this is a theoretical value and it's always a good idea to double-check your calculations before applying any voltage. Safety first!
To find the maximum voltage that can be applied across the whole network, we need to calculate the maximum power dissipation for each resistor and then find the total power dissipation.
Step 1: Calculate the maximum power dissipation for each resistor.
The formula to calculate power dissipation is:
P = V^2 / R
where P is the power dissipation in watts, V is the voltage across the resistor in volts, and R is the resistance in ohms.
For a 0.5W resistor, we can rearrange the formula to find the maximum voltage:
V = sqrt(P * R)
For the 2.5-kohms resistor:
V1 = sqrt(0.5W * (2.5-kohms)) = 1.58V
For the 3.7-kohms resistor:
V2 = sqrt(0.5W * (3.7-kohms)) = 1.92V
For the 1.4-kohms resistor:
V3 = sqrt(0.5W * (1.4-kohms)) = 1.18V
Step 2: Find the total resistance.
The total resistance of resistors connected in series is the sum of their individual resistances:
R_total = R1 + R2 + R3
R_total = 2.5-kohms + 3.7-kohms + 1.4-kohms
R_total = 7.6-kohms
Step 3: Find the maximum voltage across the whole network.
The maximum voltage across the whole network is the sum of the maximum voltages across each resistor:
V_total = V1 + V2 + V3
V_total = 1.58V + 1.92V + 1.18V
V_total = 4.68V
Therefore, the maximum voltage that can be applied across the whole network is 4.68 volts.
To find the maximum voltage that can be applied across the whole network, we need to calculate the total power dissipated by the resistors. Since the resistors are connected in series, the same current flows through all of them.
First, calculate the total resistance of the parallel combination of the 2.5-kohms and 3.7-kohms resistors:
1/R_parallel = 1/R1 + 1/R2
1/R_parallel = 1/2.5 + 1/3.7
1/R_parallel = 0.4 + 0.27
1/R_parallel = 0.67
R_parallel = 1 / 0.67
R_parallel ≈ 1.49 kohms
The total resistance of the series combination of the parallel resistors and the 1.4-kohms resistor is:
R_total = R_parallel + R3
R_total = 1.49 + 1.4
R_total = 2.89 kohms
Next, we can calculate the total current flowing through the network by using Ohm's Law:
I = V / R_total
Given that each resistor is rated for a maximum power of 0.5W, we know that the maximum current flowing through each resistor can be calculated using the formula:
I_max = sqrt(P_max / R)
I_max = sqrt(0.5 / R)
I_max = sqrt(0.5 / 2.5) ≈ 0.45 A (for the 2.5-kohms resistor)
I_max = sqrt(0.5 / 3.7) ≈ 0.36 A (for the 3.7-kohms resistor)
I_max = sqrt(0.5 / 1.4) ≈ 0.61 A (for the 1.4-kohms resistor)
Since the current in series connections is the same, we take the largest value, which is approximately 0.61 A.
Now, we can find the maximum voltage that can be applied across the network by multiplying the total current by the total resistance:
V_max = I_max * R_total
V_max = 0.61 * 2.89
V_max ≈ 1.77 V
Therefore, the maximum voltage that can be applied across the whole network without overheating the resistors is approximately 1.77 volts.
R1 = 1.4k Ohms
R2 = 2.5k Ohms
R3 = 3.7k Ohms
P1 max = I1^2*R1 = 500 mW
I1^2*1.4 = 500
I1^2 = 357.14
I1 = 18.9mA, max = Total current.
Req. = R1 + (R2*R3)/(R2+R3) =
1.4 + (2.5*3.7)/(2.5+3.7) = 1.4 + 1.49 =
2.89k Ohms
E max = I1*Req = 18.9 * 2.89 = 54.6 V. =