A circuit is a closed path through which current flows continuously, driven by an EMF source such as a battery. Kirchhoff's two laws — charge conservation at every junction, and energy conservation around every loop — give us a complete toolkit to analyze any DC network, no matter how complex.
Key Concepts
EMF and Internal Resistance
An EMF source (battery, generator) maintains a potential difference. A real battery has internal resistance r, so its terminal voltage is Vterm=ε−Ir, where ε is the EMF and I is the current drawn. Terminal voltage drops under load.
Kirchhoff's Current Law (KCL)
At any junction in a circuit, the sum of currents entering equals the sum leaving: ∑Iin=∑Iout. This is conservation of charge — charge cannot accumulate at a node.
Kirchhoff's Voltage Law (KVL)
The sum of all potential differences around any closed loop is zero: ∑ΔV=0. This is conservation of energy — a charge returning to its starting point gains and loses the same total energy.
Resistors in Series and Parallel
Series: same current through each; Req=R1+R2+⋯. Parallel: same voltage across each; 1/Req=1/R1+1/R2+⋯. The equivalent parallel resistance is always less than the smallest individual resistor.
RC Circuits
A capacitor charging through a resistor: voltage builds exponentially with time constant τ=RC. At t=τ the capacitor reaches ≈63% of full charge. After 5τ it is essentially fully charged (>99%). Discharging follows the same exponential decay.
Key Equations
Terminal voltage
Vterm=ε−Ir
Actual voltage delivered by a battery with EMF ε and internal resistance r when supplying current I.
Series resistors
Req=R1+R2+⋯
Equivalent resistance of resistors in series. Same current flows through all.
Parallel resistors
Req1=R11+R21+⋯
Equivalent resistance of resistors in parallel. Same voltage across all.
RC time constant
τ=RC
Time constant for charging/discharging. After time τ, capacitor is at 63% of final voltage; after 5τ it is >99% charged.
Capacitor charging voltage
VC(t)=ε(1−e−t/τ)
Voltage across capacitor as a function of time when charging from 0 through resistor R from EMF ε.
Worked Example
Series-Parallel Circuit with Internal Resistance
Problem
A 12 V battery with internal resistance r=1Ω is connected to R1=5Ω in series with a parallel combination of R2=6Ω and R3=3Ω. Find the total current drawn and the terminal voltage.
Solution
First reduce the parallel combination:
R23=R2+R3R2R3=6+3(6)(3)=918=2Ω
Total external resistance and full circuit resistance:
Rext=R1+R23=5+2=7ΩRtotal=r+Rext=1+7=8Ω
Total current from the battery:
I=Rtotalε=812=1.5 A
Terminal voltage (voltage at battery terminals):
Vterm=ε−Ir=12−(1.5)(1)=10.5 V
AnswerI = 1.5 A; terminal voltage = 10.5 V.
Practice
Exercises
7 problems
1of 7
Three resistors R1=4Ω, R2=6Ω, and R3=10Ω are connected in series to a V=20 V battery. What current (in A) flows through the circuit?
A
2of 7
Resistors R1=6Ω and R2=12Ω are connected in **parallel**. What is the equivalent resistance (in Ω)?
Ω
3of 7
A battery has EMF ε=9.0 V and internal resistance r=0.50Ω. When it delivers current I=2.0 A, what is the terminal voltage (in V)?
V
4of 7
Resistors R1=8Ω and R2=4Ω are in series with a 12 V battery. What is the voltage (in V) across R2?
V
5of 7
A current of I=3.0 A flows through a resistor R=8.0Ω. What power (in W) is dissipated in it?
W
6of 7
An RC circuit has R=500Ω and C=100μF. What is the time constant τ (in ms)?
ms
7of 7
A 2Ω resistor is in series with a parallel combination of two 4Ω resistors, all connected to a 12 V battery. What is the total current (in A) drawn from the battery?
A
Key Takeaways
KCL (charge conservation) at every node; KVL (energy conservation) around every loop — these two rules solve any DC circuit.
Series resistors share the same current; parallel resistors share the same voltage. Use this to spot which applies.
Terminal voltage drops under load: Vterm=ε−Ir. A battery with high internal resistance is less useful under heavy load.
RC time constant τ=RC: larger R or larger C means slower charging/discharging.
The voltage divider rule for two series resistors: VR2=Vsource⋅R2/(R1+R2).