Calculating Current and Voltage in a Series Resistor Circuit

  • Thread starter Thread starter Midas_Touch
  • Start date Start date
  • Tags Tags
    Resistors Series
AI Thread Summary
In a series resistor circuit, the current remains constant at all points, while the voltage drops across each resistor. The equivalent resistance (Req) is the sum of R1 and R2, and the total voltage drop across the resistors must equal the source voltage (Vs). Ohm's law can be applied to calculate the voltage at each point based on the constant current. If R2 were parallel to R1, the circuit would no longer be a series circuit, leading to different current and voltage behaviors. Understanding these principles is crucial for accurately analyzing circuit configurations.
Midas_Touch
|------A------|
|.....* R2
- Vs.....|B
-.....|
|.....* R1
|------C----- |

Assuming that the above is a circuit (sorry for the bad drawing, ignore the dots), what would be Ia, Ib, Ic (currents), Va, Vb, Vc (voltages)? Initially, I thought that the current and the voltage would be the same at all points.

I know that Req is equal to the sum of R1 and R2 (resistance). So would Ia be Vs/(R1*R2)? Would Ib be Vs/R2? and Ic, Vs/R1?

For the voltages, I think that it is equal to Vs at points A,B,C. But, I am not sure.

I'd appreciate some feedback. Thanks.
 
Last edited by a moderator:
Physics news on Phys.org
In a series circiut the current is the same at all points, not the voltage. Use Ohms law to find the voltage at each point, given the constant current. Your total voltage drop should equal the source voltage.
 
Interestingly you are using two correct concepts, but not applied to the correct variable.

It is the voltage (potential) that starts off at a value (Vs) and in a series circuit, drops some amount through each component, until it reaches its lowest value (typically ground = 0v).

The other variable current, in a series arrangement, does not drop but stays the same through each component.

This will be different in a parallel arrangement, and I will let you figure that one out.. Applying what I gave you above, try this one again..
 
Last edited:
Integral said:
In a series circiut the current is the same at all points, not the voltage. Use Ohms law to find the voltage at each point, given the constant current. Your total voltage drop should equal the source voltage.

How about if R2 was parallel to R1? Would the current still be the same and the voltage be different at points around the circuit?
 
Midas_Touch said:
How about if R2 was parallel to R1? Would the current still be the same and the voltage be different at points around the circuit?
See my last post, and you should be able to deduce your answer.
 
Midas_Touch said:
How about if R2 was parallel to R1? Would the current still be the same and the voltage be different at points around the circuit?
If R2 were parallel to R1 then it would not be a series circuit would it? Why do you expect a parallel circuit to behave the same as a series?
 
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 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
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