let's call :
- Rx the resistance, part of the 500 ohm resistance, connected in parallel to the load resistance RL - - Rp the parallel between Rx and RL
We can write the following equation :
200/((Rx//RL)+(500-Rx)) = 25/(RL//Rx)...which can also be written :
(200-25)/(500-Rx) = 25/((250*Rx)/(250+Rx)
175*(250*Rx)/(250+Rx) = 12500-25Rx
43750*Rx = (250+Rx)(12500-25Rx)
43750*Rx = 3,125,000-6250Rx+12500Rx-25Rx^2
(43750-6250)Rx -3,125,000+25Rx^2 = 0
37500Rx-3,125,000+25Rx^2 = 0
dividing all terms by 1000 :
-0.025Rx^2-37.5Rx+3125 = 0
Rx = (37.5±√(37.5^2+3125*0.1))/-0,05 = 79.16 ohm
Rp = Rx//RL = (250*79.16)/( (250+79.16) = 60.12 ohm
Verification :
25/Rp = 200/(500-Rx+Rp)
25/60,12 = 0.4158
200/(500-79.16+60.12) = 0.4158
calculation is perfectly right
Total power consumption :
Rt = Rp+(500-Rx) = 60.12-79.16+500 = 481 ohm
P = V^2/Rt = 40,000/481 = 83.16 watt
Two tapping points? why and how are there two? Do you mean two solutions? I only get one.
Load is R = E/I = 25/0.1 = 250 ohms
Call the tap point x. (in ohms from the bottom)
resistance from that point to common is the unknown in parallel with 250 = 250x/(250+x)
total R is the top portion, (500–x) in series with above value, or (500–x) + (250x/(250+x))
voltage at junction, 25 volts, is (voltage divider rule)
25 = 200 [ 250x/(250+x) ] / [ 500 – x + (250x/(250+x)) ]
now just solve for x
[ 250x/(250+x) ] / [ 500 – x + (250x/(250+x)) ] = 1/8
cross multiply
(500 – x) + (250x/(250+x)) = 8?250x/(250+x)
multiply by (250+x)
(500 – x)(250+x) + 250x = 2000x
simplify
–x2 + 500x – 250x + 125000 + 250x – 2000x = 0
x2 + 500x – 2000x + 125000 = 0
x2 – 1500x + 125000 = 0
x = 88.6, 1411
skipping the 1411 solution as impossible, answer is 88.6 ohms
to get power, substitute x back in
Total R is (500–x) + (250x/(250+x))
and use P = E2/R
edit, corrected.
The voltage tapped offis 1/8 of the supply so it is 1/8 of 500 ohms or 62.5 ohms from the bottom.
and 500-62.5 = 437.5 ohms from the top. The current is 200/500 = .4A in the 62.5 ohm resistance but 0.5 A in the 437.5 ohm resistance so the power is 25*.4+ 437.5*.5= 228.75 W
Let R1 = top portion of 500 Ohm resistor
Let R2 = bottom portion of 500 Ohm resistor = (500 Ohm - R1)
I of R1 = 175V/R1 = 25V/R2 + .1A
So,
175/R1 = 25/(500 - R1)
(175/R1) - [25/(500 - R1)] = 0
.1R1^2 - 250R1 + 87,500 = 0
R1 = 420.844 or 2079.156 Ohms but R1 must be less than 500 Ohms so R1 = 420.844 Ohms
R2 = (500 - 420.844)Ohms = 79.156 Ohms
R2 ll RL = 79.156 ll 250 = 60.12 Ohms
Total power consumed = V^2/Total resistance =[(200V^2) / (420.844 + 60.12)Ohm = 83.17 Watts
The principles known as Kirchhoff's rules, in honor of the man who developed them, provide a means of obtaining enough independent equations to solve for the currents flowing in an electrical circuit. There are many ways of stating Kirchhoff's rules but this experiment uses the formalism described below.
Kirchhoff's junction rule states that the algebraic sum of the currents at any branch point or junction in a circuit is zero. Symbolically, we may write Equation 1, below, as the sum of n currents flowing into a junction:
(Eq. 1)
In applying this rule, currents flowing into a branch point are taken as positive, while currents flowing out of the branch point are taken as negative. the opposite sign convention could also be used, but this would simply multiply the equation by (-1) and therefore would provide exactly the same information.
Kirchhoff's second rule states that the algebraic sum of the potential changes around any complete loop in the network is zero. Again we may write Eq. 2, symbolically ,as the sum of the potential changes:
(Eq. 2)
AT 250 OHM POINT.
A voltage of 200 volts is applied to tapped resistor of 500 ohms. Find the resistance between two tapping points connected to a circuit needing 0.1 ampere at 25 volts. Calculate the total power consumed?
