Kirchhoff's Rule / finding unknown resistances and voltages

  • Thread starter Thread starter exi
  • Start date Start date
AI Thread Summary
The discussion focuses on applying Kirchhoff's Rule and Ohm's Law to find the current through resistor R1 (2 Ω) and the voltage across the adjacent cell. The user initially considers mesh analysis with an assumed current of 3 A and aims to calculate the unknown current I(2) and the voltage of the cell. They suggest combining resistors R2 (6.5 Ω) and a 4 Ω resistor to simplify the circuit analysis, grounding the right side at the negative terminal of a 24V source. The user realizes the importance of correctly combining voltage and current-resistance products in their calculations. Ultimately, they successfully solve the problem by applying these concepts.
exi
Messages
85
Reaction score
0

Homework Statement



Find the current passing through R1 and the voltage passing through the cell to the immediate left of R1.

R1 is 2 Ω, and R2 is 6.5 Ω.

http://img255.imageshack.us/img255/9520/problem2ux3.png

Homework Equations



Kirchhoff's Rule; Ohm's Law

The Attempt at a Solution



Not sure if I'm approaching this the correct way. What I had in mind was to do some mesh analysis while considering I(1) to be 3 A, doing Kirchhoff's for the bottom half of the circuit, finding I(2), and using Ohm's to find I through that 2 Ω resistor.

Little unsure about finding the voltage of that mystery cell, though.
 
Last edited by a moderator:
Physics news on Phys.org
I'd first combine R2 and the 4 Ohm resistor -- no need to keep them separate for this problem. Then ground the right side, at the - side of the 24V voltage source. You then have 2 unknown node voltages that you can write the KCL equations for, and once you solve for them, you have the solutions for the question.

BTW, instead of saying "the voltage passing through the cell to the immediate left of R1", it would be better to say "the voltage across the cell". Current passes through an element in response to the voltage placed across the element.
 
berkeman said:
I'd first combine R2 and the 4 Ohm resistor -- no need to keep them separate for this problem. Then ground the right side, at the - side of the 24V voltage source. You then have 2 unknown node voltages that you can write the KCL equations for, and once you solve for them, you have the solutions for the question.

BTW, instead of saying "the voltage passing through the cell to the immediate left of R1", it would be better to say "the voltage across the cell". Current passes through an element in response to the voltage placed across the element.

Well, that worked beautifully. I kept thinking I couldn't combine terms in the first part of the Kirchhoff's work for some reason; took me a second to realize that a voltage and the product of a current and a resistance value (Ohm's, anyone?) definitely are combinable.

My "duurrrrr" moment for the day, I suppose.

Thanks.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top