B.5 · IB Physics
Kirchhoff's laws
Two cells and three resistors form a two-loop network. The junction rule and the loop rule give three equations for the three branch currents. A negative answer means the current flows against the arrow you assumed.
Predict first
Start from the default settings (ε₁ = 6.0 V, ε₂ = 3.0 V, R₁ = 2.0 Ω, R₂ = 4.0 Ω, R₃ = 6.0 Ω). If you tick Reverse cell 2, what happens to the current I₃ in R₃?
With cell 2 reversed its emf pulls the potential of P down, so V_P falls from 4.09 V to 2.45 V. Since I₃ = V_P/R₃, the current in R₃ falls from 0.682 A to 0.409 A.
Check it in the model: Tick Reverse cell 2 and compare the I₃ (in R₃) and Potential of P readouts before and after.
dots: actual conventional current (speed grows with |I|) · hollow arrows: assumed directions · Q is taken as 0 V
Presets
junction P: I₁ + I₂ = I₃
loop 1: ε₁ = I₁R₁ + I₃R₃
loop 2: ±ε₂ = I₂R₂ + I₃R₃
outer loop: ε₁ ∓ ε₂ = I₁R₁ − I₂R₂
(− if cell 2 reversed: ε₂ term negative)
Try this
- Press “Cell 2 being charged”. Which way does the current flow through cell 2 and why? Check that the potential at P is higher than ε₂.
- Press “Cells opposed: I₃ = 0” and then change R₃. Why does the current in R₃ stay zero, and what is the current round the outer loop?
- Select Challenge 1 and adjust ε₂ until I₂ becomes zero. What value of ε₂ does that need, and how does it relate to the potential of P?
Common error: changing the sign of a current halfway through a solution; if I₂ comes out as −0.5 A, keep −0.5 A in every later equation and only at the end read it as 0.5 A flowing against the assumed arrow.