C#•2y ago

❔ Using a loop I need to increment voltage by 0.1.

At each increment the power dissipated in R2 (P2) should be calculated (should be able to use P2 = V2 * I2). The voltage loop should stop when P2 = 1.5. Having a hard time figuring out how to get P2 to be calculated at each increment since V2 and I2 would be needed and those would need to calculated first.

14 Replies

MattKOP•2y ago

90% sure that code is rubbish

MattKOP•2y ago

this is another attempt

Doombox•2y ago

why do while?

for(decimal v = 0; v < 1.5m; v += 0.1m)

for(decimal v = 0; v < 1.5m; v += 0.1m)

would even suffice if you simply need to increase the voltage by 0.1 per increment, then in the loop you can do whatever you need to do what are you actually trying to calculate? can't figure it out looks like ohms law to me, but to what end

MattKOP•2y ago

Trying to calculate power( P = VI ). Let me grab a photo of the circuit for clarification

MattKOP•2y ago

https://gyazo.com/f86be9d0a7c627ffc13784132288954d

Gyazo

MattKOP•2y ago

essentially I'm trying to calculate what source voltage for V1 would deliver 1.5 watts to R2 the loop itself should terminate after it reaches that point

Doombox•2y ago

ah good old resistor circuits, got 2 in series with the 1 in parallel, you don't necessarily need to brute force this with software honestly though I haven't done electronics in a while, you should be able to trial and error this one pretty quick uhh would 25 volts do it? not sure if my brain is awake enough to remember electronics stuff I barely paid attention to in the first place pepelaff

MattKOP•2y ago

haha

Doombox•2y ago

not sure what source voltage you would need, can't quite recall how parallel resistors affected voltage

Doombox•2y ago

if memory serves, resistors in parallel doesn't change the voltage across the terminals only get a voltage divider effect from series resistors

Anton•2y ago

use an infinite loop (while(true)) and check the conditions and break manually

ringsig•2y ago

series resistance
= R_1 + R_2
-> R_s

split voltage into ratio R1:R2 and get V_2
V_2 = V * R_2 / (R_1 + R_2)

power from voltage and resistance
P = V^2 / R

write P_2 in terms of resistances and V using expression for V_2
P_2
= V_2^2 / R_2
= (V * R_2 / (R_1 + R_2))^2 / R_2
= (V * R_2)^2 / (R_2 * (R_1 + R_2)^2)
= V^2 * R_2^2 / (R_2 * (R_1 + R_2)^2)
= V^2 * R_2 / (R_1 + R_2)^2

V^2 = P_2 * (R_1 + R_2)^2 / R_2

V = sqrt(P_2 * (R_1 + R_2)^2 / R_2)

P_2 = 1.5

V = sqrt(1.5 * (R_1 + R_2)^2 / R_2)

with specified values:
R_1 = 100
R_2 = 150
R_3 = 150

V
= sqrt(1.5 * (100 + 150)^2 / 150)
= sqrt(1.5 * 250^2 / 150)
= sqrt(1.5 * 62500 / 150)
= sqrt(625)
= 25

series resistance
= R_1 + R_2
-> R_s

split voltage into ratio R1:R2 and get V_2
V_2 = V * R_2 / (R_1 + R_2)

power from voltage and resistance
P = V^2 / R

write P_2 in terms of resistances and V using expression for V_2
P_2
= V_2^2 / R_2
= (V * R_2 / (R_1 + R_2))^2 / R_2
= (V * R_2)^2 / (R_2 * (R_1 + R_2)^2)
= V^2 * R_2^2 / (R_2 * (R_1 + R_2)^2)
= V^2 * R_2 / (R_1 + R_2)^2

V^2 = P_2 * (R_1 + R_2)^2 / R_2

V = sqrt(P_2 * (R_1 + R_2)^2 / R_2)

P_2 = 1.5

V = sqrt(1.5 * (R_1 + R_2)^2 / R_2)

with specified values:
R_1 = 100
R_2 = 150
R_3 = 150

V
= sqrt(1.5 * (100 + 150)^2 / 150)
= sqrt(1.5 * 250^2 / 150)
= sqrt(1.5 * 62500 / 150)
= sqrt(625)
= 25

If you need the voltage in an increment of 0.1 rounding up: ceil(V * 10) / 10

ringsig•2y ago

On paper

Accord•2y ago

Was this issue resolved? If so, run /close - otherwise I will mark this as stale and this post will be archived until there is new activity.

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❔ Using a loop I need to increment voltage by 0.1.