Respuesta :
We found a similar problem:
A 2.8-kΩ and a 2.1-kΩ resistor are connected in parallel; this combination is connected in series with a 1.8-kΩ resistor. If each resistor is rated at 1/2 W (maximum without overheating), what is the maximum voltage that can be applied across the whole network?
Answer:
The maximum voltage is 57 V
Explanation:
Circuits
The power dissipation of a resistor R and voltage V is:
[tex]\displaystyle P=\frac{V^2}{R}[/tex]
The maximum power for each resistor must not exceed 0.5 W, this allows us to find the maximum voltage for each one.
[tex]\displaystyle P=\frac{V^2}{R} =>V=\sqrt{PR}[/tex]
For R1=2800
[tex]\displaystyle V_1=\sqrt{0.5\times 2800}=37.4\ V[/tex]
For R2=2100
[tex]\displaystyle V_2=\sqrt{0.5\times 2100}=32.4\ V[/tex]
Since this voltage is less than the first one and they are connected in parallel, we choose this one for both because the greater voltage would cause overheating in the second resistor.
The voltage for R3=1200 is
[tex]\displaystyle V_3=\sqrt{0.5\times 1200}=24.5\ V[/tex]
This maximum voltage will add to the combination of the parallel in the final series connection, thus the total maximum voltage is
[tex]V_m=32.4\ V+24.5\ V=56.9\ V\approx 57\ V[/tex]
The maximum voltage is 57 V
The maximum voltage that can be applied to the whole network is 62.4V which is explained below.
Power and Voltage:
The relation between power, resistance and voltage is given by:
P = V²/R
where P is the power
R is the resistance, and
V is the voltage
It is given that the maximum power output of the resistors is 0.5W
V = √PR
so, for the 2.8kΩ resistor, the maximum voltage will be:
V = √(0.5×2800)
V = 37.4V
Similarly for the 2.1kΩ resistor, the maximum voltage will be:
V = √(0.5×2100)
V = 32.4V
which is smaller than the maximum voltage for the 2.8kΩ resistor
Since the 2.8kΩ and the 2.1kΩ resistors are connected in parallel, the voltage across them must be the same.
The maximum allowed voltage across this parallel combination will be 32.4V because if we chose 37.4 V then it will exceed the power rating of the 2.1kΩ resistor.
Now, similarly, the voltage drop across the 1.8kΩ resistor in series is:
V = √PR
V = √0.5×1800
V = 30V
So the maximum voltage that can be applied to the whole network is
32.4V + 30V = 62.4V
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