How Do Kirchhoff's Laws Apply to a Capacitor in a Circuit?

  • Thread starter Thread starter Saraphim
  • Start date Start date
  • Tags Tags
    Capacitance Law
AI Thread Summary
Kirchhoff's laws are applied to analyze a circuit with a capacitor, leading to the formulation of a differential equation for the charge on the capacitor. The node law indicates that the current through the capacitor branch is related to the currents in other branches. The discussion reveals confusion about identifying independent meshes within the circuit, with three mesh equations derived but one deemed non-independent. A suggestion is made to omit the redundant mesh equation and focus on the remaining two to express the currents in terms of the charge. The conversation emphasizes the need for clarity in selecting independent equations to solve the problem effectively.
Saraphim
Messages
46
Reaction score
0

Homework Statement


The problem considers the diagram in the photo that I have attached. Apologies for the poor quality, but sketching it on my computer would have been even worse!

a) Write down Kirchhoff's laws for the circuit and show that when the switch is closed, the charge on the capacitor Q behaves according to:
\frac{R_2}{R_1+R_2}\epsilon = \frac{Q}{C}+\left(R_3+\frac{R_1 R_2}{R_1+R_2}\right)\frac{dQ}{dt}

Homework Equations


The current through the branch with the capacitor, I_3, can be described by dQ/dt.

The Attempt at a Solution



First, Kirchhoff's laws.

The node law says that I_1=I_2+I_3.

Then the mesh laws, I get three meshes:

\epsilon=I_1 R_1 + I_2 R_2
\epsilon=I_1 R_1 + I_3 R_3+ \frac{Q}{C}
0=I_3 R_3 + \frac{Q}{C} - I_2 R_2

I'm fairly sure those are correct, however, I can't figure out how to combine these to show the differential equation given in the problem statement. I think I must leave out one of the relations obtained from the mesh law, but I have no idea where to start, or how to proceed in any sort of methodical way. Any pointers would be greatly appreciated, I'm sure I can work it out if I just get a nudge in the right direction!
 

Attachments

  • diag.jpg
    diag.jpg
    6.3 KB · Views: 379
Physics news on Phys.org
Saraphim said:



\epsilon=I_1 R_1 + I_2 R_2
\epsilon=I_1 R_1 + I_3 R_3+ \frac{Q}{C}
0=I_3 R_3 + \frac{Q}{C} - I_2 R_2



The circuit contains two independent meshes. Omit the third mesh equation, it is not independent from the previous ones. Use the equation I1=I2+I3 and the first mesh equation to get I1 in terms of I3. Substitute for I1 in the second mesh equation. Use dQ/dt for I3.

ehild
 
So the first independent mesh is the one containing R1 and R3 and the other containing R3 and R2?
 
Wait, no, that's wrong. I'll think on it some more.
 
I can't seem to find out which two meshes are independent. :rolleyes:
 
Any two ones. ehild
 
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