Week 3 · Chapter 1 (C): Kirchhoff’s first law (conservation of charge, the node rule) and second law (conservation of energy, the loop rule), the guidelines for assumed current directions and signs, and three worked multi-loop circuits · 3 lessons
2
Kirchhoff’s second law and the sign rules
Physics Sec 3 Term 1, Chapter 1: pp. 18–23
GoalWrite the loop equation of a single loop with the sign rules, and say what a negative current means.
New words
closed loop · a path through the circuit that comes back to its starting pointKirchhoff’s second law · round any closed loop the sum of the emfs equals the sum of the voltages IRassumed direction · the direction you choose for a current before solving; a negative answer reverses it
Explanation
The second law is about energy. Take one charge once round a closed loop: it returns to where it started, so the energy it gains from the sources must equal the energy it gives to the resistors. Per coulomb this reads Σ E = Σ IR, where E is the emf of a source (the VB of Week 2). To use the law, follow four rules. (1) Give every branch an assumed current direction. (2) Choose one direction round each loop. (3) A term IR is positive when the assumed current runs the same way as the loop direction, and negative when it runs against it. (4) An emf is positive when the loop crosses the battery from its negative pole to its positive pole, and negative when it crosses from positive to negative. A different choice of directions only changes the signs inside the equation: the real answer does not change.
What the loop meets
Term in the equation
a battery crossed from its − pole to its + pole
+E
a battery crossed from its + pole to its − pole
−E
a resistor, assumed current along the loop direction
+IR
a resistor, assumed current against the loop direction
−IR
Worked examples
Element
How it is met
Term
battery 10 V
crossed from − to +
+10
battery 4 V
crossed from + to −
−4
resistance 3 Ω
current I along the loop
+3I
One loop has two batteries that push against each other: E₁ = 10 V and E₂ = 4 V, and the resistance of the loop is 3 Ω. Take the loop direction and the assumed current I the same way, with the loop crossing E₁ from − to + and E₂ from + to −. Then 10 − 4 = 3I, so I = 2 A. The answer is positive, so the current really flows in the assumed direction.
Now run the loop the opposite way round the same circuit, keeping the assumed current I as before. The loop crosses E₁ from + to − (−10) and E₂ from − to + (+4), and the current runs against the loop, so the resistor term is −3I. Then −10 + 4 = −3I, and I = 2 A again: the choices changed the signs but not the result.
A loop contains a 12 V battery crossed from − to +, resistors of 2 Ω and 4 Ω, and a second battery with unknown emf E crossed from + to −. The current is 1 A along the loop. Then 12 − E = 1 × (2 + 4), so E = 6 V.
Class activity
Paper drill. Draw a rectangular loop with a 9 V battery on its left side (positive pole at the top) and a 5 Ω resistor on its right side. Go round clockwise (up the left side) and assume a clockwise current. Write the sign of each term and the equation, and solve it. Then turn the battery round so that its positive pole is at the bottom, write the equation again and say what the sign of the answer tells you.
Practice
1
One loop has a battery of 18 V crossed from − to + and resistors of 4 Ω and 5 Ω. The assumed current runs along the loop direction. Find I.
18 = I × (4 + 5), so I = 2 A.
2
Two batteries oppose each other in one loop: the loop crosses 15 V from − to + and 6 V from + to −. The loop resistance is 3 Ω and the assumed current runs along the loop. Find I.
15 − 6 = 3I, so I = 3 A.
3
In the previous question the loop is run the opposite way round while the assumed current keeps its direction. Write the equation and the result.
The emfs change sign and the IR term is against the loop: −15 + 6 = −3I, so I = 3 A, the same answer.
4
A loop crosses a 20 V battery and an 8 V battery, both from − to +, and the loop resistance is 7 Ω. The assumed current runs along the loop. Find I.
