1. Resistors & Capacitors in Series and Parallel
Kirchoff’s rules

2. Circuit diagrams
Why diagrams? Because drawing batteries, light bulbs, resistors, etc in detail takes a loooooooong time
Some basic symbols

3. Series circuits Everything in one loop
Only one path for current to take, so all charges must move at the same rate. Current is the same everywhere in the circuit.
Each element in a circuit uses some energy to do whatever it does
Light bulbs produce light (and heat), resistors produce heat
Voltage is a measure of the energy in the circuit. Energy is conserved, so the total voltage supplied by the battery must be used completely by the elements in the circuit.
𝐼= 𝐼 1 = 𝐼 2 = 𝐼 3 =…
𝑉= 𝑉 1 + 𝑉 2 + 𝑉 3 +…

4. Resistors in Series circuits
Each resistor is like another hurdle for the current to jump over, so resistors in series increase the resistance, as we saw in our experiment.
𝑅 𝐸𝑄 = 𝑅 1 + 𝑅 2 + 𝑅 3 +…
Example: If 𝑅 1 =3Ω, 𝑅 2 =7Ω 𝑅 3 =30Ω, determine the equivalent resistance of the circuit.
𝑅 𝐸𝑄 =3Ω+7Ω+30Ω=40Ω

5. Capacitors in Series
As seen in the video, the positive plate on one capacitor is connected to the negative side on the next capacitor, so they all have the same charge, but different voltages.
1 𝐶 𝐸𝑄 = 1 𝐶 𝐶 𝐶 3 +…
Example: if 𝐶 1 =10𝐹, 𝐶 2 =15𝐹 𝐶 3 =30𝐹, determine the equivalent capacitance of the circuit.
1 𝐶 𝐸𝑄 = = = 6 30 = 1 5
𝐶 𝐸𝑄 =5 𝐹

6. Parallel circuits Everything connects to the battery
The current has an option of paths to take, so the current in each branch or element could be different, charge is conserved, so the current in each branch should add at any junction.
Voltage is a measure of the energy in the circuit. Energy is conserved, so the total voltage supplied by the battery must be used completely by the elements in the circuit.
But each element has its own direct path to the battery, so each element gets the same amount of voltage across it as the battery can provide.
𝐼= 𝐼 1 + 𝐼 2 + 𝐼 3 +…
𝐼= 𝑉 1 = 𝑉 2 = 𝑉 3 =…

7. Resistors in Parallel circuits
Each resistor is connected to the battery and has the same voltage across it as the battery, but each resistor could have a different current, and therefore, the equivalent resistance for ressitors in parallel is
1 𝑅 𝐸𝑄 = 1 𝑅 𝑅 𝑅 3 +…
Example: Find the equivalent resistance for the circuit above.
1 𝑅 𝐸𝑄 = = = =0.543
𝑅 𝐸𝑄 =1.8 Ω

8. Capacitors in Parallel
The positive plates are all connected to the positive terminal of the battery, giving each capacitor the same voltage, but different amounts of charge.
𝐶 𝐸𝑄 = 𝐶 1 + 𝐶 2 + 𝐶 3 +…
Example: Determine the equivalent capacitance for the circuit above.
𝐶 𝐸𝑄 =0.1𝜇𝐹+0.2𝜇𝐹+0.3𝜇𝐹=0.6𝜇𝐹

9. Example
Determine the equivalent resistance for the circuit shown assuming each resistor is 40 Ω.
Parallel: 1 𝑅 𝑃 = = 2 40 = 1 20 ; 𝑅 𝑃 =20Ω
Series: 𝑅 𝐸𝑄 =40Ω+40Ω+20Ω= 100Ω

10. Kirchoff’s Rules for circuits
The junction rule: current in = current out
“The principle of conservation of charge implies that at any node (junction) in an electrical circuit, the sum of currents flowing into that node is equal to the sum of currents flowing out of that node.”
The loop rule: the sum of emfs = the sum of the voltage drops
“The principle of conservation of energy implies that the directed sum of the electrical potential differences (voltage) around any closed network is zero.”

11. Summary
Kirchoff’s rules guide how current, voltage, and various elements like resistors and capacitors act in series and parallel configurations.
We can use these rules to determine current, voltage, capacitance & resistance in a circuit.

12. Solving Problems using Kirchoff’s Rules
Label current directions – if you get a negative answer, it just means it’s the opposite direction than you predicted.
Label your loops & move around it.
moving around the circuit in the direction of the current
Crossing resistors is a voltage drop
Crossing batteries from negative to positive adds voltage
moving around the circuit opposite the direction of the current
Crossing resistors is a voltage gain
Crossing batteries from negative to positive loses voltage

13. Example According to the junction rule: 𝐼 1 − 𝐼 2 − 𝐼 3 =0
According to the Loop rule:
− 𝑅 3 𝐼 3 + ℰ 1 − 𝑅 1 𝐼 1 =0
and
− 𝑅 3 𝐼 3 − ℰ 2 − ℰ 1 − 𝑅 2 𝐼 2 =0
And we get a system of equations that we can use to solve for one thing and plug it into the other equations.

14. Example
And we get a system of equations that we can use to solve for one thing and plug it into the other equations.
𝐼 1 − 𝐼 2 − 𝐼 3 =0
− 𝑅 3 𝐼 3 + ℰ 1 − 𝑅 1 𝐼 1 =0
− 𝑅 3 𝐼 3 − ℰ 2 − ℰ 1 − 𝑅 2 𝐼 2 =0
Assuming 𝑅 1 =100Ω, 𝑅 2 =200Ω, 𝑅 3 = 300Ω, ℰ 1 =3 𝑉, ℰ 2 =4 𝑉, we can use the equations to solve for the currents.
𝐼 1 = 𝑎𝑚𝑝𝑠, 𝐼 2 = 𝑎𝑚𝑝𝑠, 𝐼 3 =− 𝑎𝑚𝑝𝑠