Find Power in 7k Resistor from Potential Divider

[luigihs]

With Vout not connected to any additional circuitry, what power is dissipated in the 7 k resistor?

So I am trying to get the power in 7k resistor I know the formula if I need to get Vout but in the question it says that the vout is not connected to any additional circuitry so I assume that I don't have to use the Vout so is Vin / resistor in this case 10 / 7 = 1.42 <-- but this is the resistance how I am suppose to get the power?

Im assuming this from the potential divider I = Vin / R1 + R2 . Now Ohms laws say that V = I x R so if I get the total resistance between 7 + 3 = 10 divided by the voltage 10 / 10 = 1 but I still stuck because I am getting the total voltage I want only the power in 7k I am confused help me thanks!

 

[SunnyBoyNY]

Method A)

Calculate branch current : I = Vin / Rtotal. Current in R1 and R2 is the same. Power dissipated in each resistor is Px = Rx * I^2

Method B)

Calculate voltage drop across R1 or R2 using the voltage divider formula.

VR2 = Vin * R2/(R1+R2).

Power dissipated in resistor R2 is P2 = V2^2/R2.

These two methods give identical results.

Vout not connected simply means that no current flows from the Vout node to an external circuitry. In practice, you would use such circuit to scale input voltage to a CMOS chip with ~1uA input current. Voltage Vout would then be a little bit lower.

 

[tiny-tim]

hi luigihs!

... is Vin / resistor in this case 10 / 7 = 1.42 <--

what's "Vin"?

you need to use the voltage across the 7Ω resistor

Im assuming this from the potential divider I = Vin / R1 + R2 . Now Ohms laws say that V = I x R so if I get the total resistance between 7 + 3 = 10 divided by the voltage 10 / 10 = 1 …

what is this supposed to be?

 

[luigihs]

Ok so is series , 7 +3 = 10 I = 10/10 = 1 then Px =7 x 1^2 = 7 ? sounds weird is this the right answer ??

 

[SunnyBoyNY]

Ok so is series , 7 +3 = 10 I = 10/10 = 1 then Px =7 x 1^2 = 7 ? sounds weird is this the right answer ??

Yes. The total power supplied by the supply is :

I = 10 / (7k + 3k) = 1 mA.

Psupply = Vin * I = 10 * 1mA = 10 mW.

P1 = 7 kOhm *1 mA*1 mA = 7e3 * 1e-6 W = 7e-3 W = 7 mW.
P2 = 3 mW.

Psupply = P1 + P2 (Power Conservation)

Voltage, current, and power quantities might look confusing provided one of them is 1, 10, 100, and so on. The other two quantities will then have either the same or reciprocal magnitude (disregarding the decimal point). Say, R = 1, V = 20 V -> I = 20 A.

 

[luigihs]

P1 = 7 kOhm *1 mA*1 mA = 7e3 * 1e-6 W = 7e-3 W = 7 mW. <-- wait this confused is not just 7 x 1^2 or why you using 7e-3 ??

 

[SunnyBoyNY]

P1 = 7 kOhm *1 mA*1 mA = 7e3 * 1e-6 W = 7e-3 W = 7 mW. <-- wait this confused is not just 7 x 1^2 or why you using 7e-3 ??

I would suggest to take a pen and a piece of paper and go through the calculations yourself. It will be a great investment for the future.