Determine the resistance values for a voltage divider that must meet the following specifications:_______.
1 answer:
Answer:
i) when circuit is unloaded : R1 + R2 = 2kΩ.
ii) when 5V output voltage is applied : R1 = 1 kΩ , R2 = 1 kΩ
iii) when 2.5 v output voltage is applied : R1 = 1500 Ω, R2 = 500 Ω
iv) when: R1 = 1 kΩ , R2 = 1 kΩ is connected in parallel output voltage < 5 V
When : R1 = 1500 Ω, R2 = 500 Ω is connected in parallel output voltage > 2.5V
Explanation:
Current drawn from source under loaded condition ≤ 5 mA
source voltage = 10 v , required output = 5 v , 2.5 v
attached below is the sketch of the circuit
Resistance values
<em>i) when the circuit is unloaded </em>
Req = R1 + R2 = 2 kΩ ( Req = Vs / I = 10 / 5*10^-3 = 2 kΩ )
<em>ii) when output voltage = 5 v</em>
we will apply voltage divider rule
R1 = 1 kΩ ,
R2 = 1 kΩ
<em>iii) When the output voltage = 2.5 v </em>
applying voltage divider rule
R1 = 1500 Ω
R2 = 500 Ω
iv) when the load is connected to each tap one at a time
i.e. when the resistance are in parallel
when: R1 = 1 kΩ , R2 = 1 kΩ is connected in parallel output voltage < 5 V
When : R1 = 1500 Ω, R2 = 500 Ω is connected in parallel output voltage > 2.5V
attached below is the detailed solution to the given problem
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Answer:
Explanation:
Given that :
The initial volume of water = 1000.00 mL = 1000000 mm³
The initial temperature of the water = 10° C
The height of the water column h = 1000.00 mm
The final temperature of the water = 70° C
The coefficient of thermal expansion for the glass is ∝ =
The objective is to determine the the depth of the water column
In order to do that we will need to determine the volume of the water.
We obtain the data for physical properties of water at standard sea level atmospheric from pressure tables; So:
At temperature the density of the water is
At temperature the density of the water is
The mass of the water is
Thus; we can say ;
⇒
Thus, the volume of the water after heating to a required temperature of is 1022400 mm³
However; taking an integral look at this process; the volume of the water before heating can be deduced by the relation:
The area of the water before heating is:
The area of the heated water is :
Finally, the depth of the heated hot water is:
Hence the depth of the heated hot water is