Resistance Circuit Homework: Solving for Total Resistance

  • Thread starter Thread starter alfredo24pr
  • Start date Start date
  • Tags Tags
    Circuit Resistance
AI Thread Summary
The discussion focuses on solving a resistance circuit problem involving resistors in both parallel and series configurations. Participants emphasize the importance of calculating total resistance by first determining the effective resistance of R2 and R3 in parallel, then combining that with R1, and finally adding R4 in series. There's a suggestion to estimate an initial voltage to find the voltage drop across R4, which helps in calculating the current through the inner resistors. Additionally, the relationship between the current through R2 and the voltage across it is highlighted as crucial for understanding the circuit dynamics. The conversation underscores the need for careful application of Ohm's Law and the rules for combining resistances.
alfredo24pr
Messages
49
Reaction score
0

Homework Statement



http://i1103.photobucket.com/albums/g480/Alfredo24PR/physicssucks.jpg"

Homework Equations



V=IR

The Attempt at a Solution



I tried adding the resistance:

1/8 + 1/ (6+5+10) = 1/8 + 1/21 = 0.1726

1/0.1726 = 5.793 ohm
 
Last edited by a moderator:
Physics news on Phys.org
You could start out by determining the resistance of the whole circuit. R2 and R3 are in parallel. You can find a resistance for them. Together, they are in parallel with R1. You can determine this resistance. This is in series with R4. The result is the total resistance of the circuit.

Keep in mind the sum of the currents passing through R1, R2, R3 must equal the current passing through R4. Be careful how you compute effective resistance for resistors in parallel and series.
 
LawrenceC said:
You could start out by determining the resistance of the whole circuit. R2 and R3 are in parallel. You can find a resistance for them. Together, they are in parallel with R1. You can determine this resistance. This is in series with R4. The result is the total resistance of the circuit.

Keep in mind the sum of the currents passing through R1, R2, R3 must equal the current passing through R4. Be careful how you compute effective resistance for resistors in parallel and series.

Ok, I figured out the middle ones and you explained the last one.

How do I start with the first question: "What potential difference must be applied to the external terminals?"
 
One way to go at it would be to guess a voltage over the whole system. This allows you to compute the voltage drop through R4. Subtract that from the applied guessed voltage and you get the voltage drop across the inner part of the circuit. Knowing that you can compute the current through the 'inside' resistors. Keep reguessing until you get 4 amps through R2.
 
alfredo24pr said:
I tried adding the resistance:

1/8 + 1/ (6+5+10) = 1/8 + 1/21 = 0.1726

1/0.1726 = 5.793 ohm

Your formula doesn't look right. Note that while R1, R2, and R3 are in parallel, R4 is in series with that parallel group.

Here's a big hint :smile:. If you're given the current through resistor R2 then you automatically know the voltage across it (Ohm's Law). What do you know about the voltage across parallel components?
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top