← General Physics II
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DC Circuits

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 rr, so its terminal voltage is Vterm=εIrV_{\text{term}} = \varepsilon - Ir, where ε\varepsilon is the EMF and II 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\sum I_{\text{in}} = \sum I_{\text{out}}. 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\sum \Delta V = 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+R_{\text{eq}} = R_1 + R_2 + \cdots. Parallel: same voltage across each; 1/Req=1/R1+1/R2+1/R_{\text{eq}} = 1/R_1 + 1/R_2 + \cdots. 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\tau = RC. At t=τt = \tau the capacitor reaches 63%\approx 63\% of full charge. After 5τ5\tau it is essentially fully charged (>99%>99\%). Discharging follows the same exponential decay.

Key Equations

Terminal voltage
Vterm=εIrV_{\text{term}} = \varepsilon - Ir
Actual voltage delivered by a battery with EMF ε and internal resistance r when supplying current I.
Series resistors
Req=R1+R2+R_{\text{eq}} = R_1 + R_2 + \cdots
Equivalent resistance of resistors in series. Same current flows through all.
Parallel resistors
1Req=1R1+1R2+\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots
Equivalent resistance of resistors in parallel. Same voltage across all.
RC time constant
τ=RC\tau = 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(1et/τ)V_C(t) = \varepsilon\bigl(1 - e^{-t/\tau}\bigr)
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 1212 V battery with internal resistance r=1Ωr = 1\,\Omega is connected to R1=5ΩR_1 = 5\,\Omega in series with a parallel combination of R2=6ΩR_2 = 6\,\Omega and R3=3ΩR_3 = 3\,\Omega. Find the total current drawn and the terminal voltage.

Solution

First reduce the parallel combination:

R23=R2R3R2+R3=(6)(3)6+3=189=2ΩR_{23} = \frac{R_2 R_3}{R_2 + R_3} = \frac{(6)(3)}{6+3} = \frac{18}{9} = 2\,\Omega

Total external resistance and full circuit resistance:

Rext=R1+R23=5+2=7ΩRtotal=r+Rext=1+7=8ΩR_{\text{ext}} = R_1 + R_{23} = 5 + 2 = 7\,\Omega \qquad R_{\text{total}} = r + R_{\text{ext}} = 1 + 7 = 8\,\Omega

Total current from the battery:

I=εRtotal=128=1.5 AI = \frac{\varepsilon}{R_{\text{total}}} = \frac{12}{8} = 1.5\text{ A}

Terminal voltage (voltage at battery terminals):

Vterm=εIr=12(1.5)(1)=10.5 VV_{\text{term}} = \varepsilon - Ir = 12 - (1.5)(1) = 10.5\text{ V}
Answer I = 1.5 A; terminal voltage = 10.5 V.
Practice

Exercises

7 problems
1 of 7

Three resistors R1=4ΩR_1 = 4\,\Omega, R2=6ΩR_2 = 6\,\Omega, and R3=10ΩR_3 = 10\,\Omega are connected in series to a V=20V = 20 V battery. What current (in A) flows through the circuit?

A
2 of 7

Resistors R1=6ΩR_1 = 6\,\Omega and R2=12ΩR_2 = 12\,\Omega are connected in **parallel**. What is the equivalent resistance (in Ω)?

Ω
3 of 7

A battery has EMF ε=9.0\varepsilon = 9.0 V and internal resistance r=0.50Ωr = 0.50\,\Omega. When it delivers current I=2.0I = 2.0 A, what is the terminal voltage (in V)?

V
4 of 7

Resistors R1=8ΩR_1 = 8\,\Omega and R2=4ΩR_2 = 4\,\Omega are in series with a 1212 V battery. What is the voltage (in V) across R2R_2?

V
5 of 7

A current of I=3.0I = 3.0 A flows through a resistor R=8.0ΩR = 8.0\,\Omega. What power (in W) is dissipated in it?

W
6 of 7

An RC circuit has R=500ΩR = 500\,\Omega and C=100μC = 100\,\muF. What is the time constant τ\tau (in ms)?

ms
7 of 7

A 2Ω2\,\Omega resistor is in series with a parallel combination of two 4Ω4\,\Omega resistors, all connected to a 1212 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=εIrV_{\text{term}} = \varepsilon - Ir. A battery with high internal resistance is less useful under heavy load.
  • RC time constant τ=RC\tau = RC: larger RR or larger CC means slower charging/discharging.
  • The voltage divider rule for two series resistors: VR2=VsourceR2/(R1+R2)V_{R_2} = V_{\text{source}} \cdot R_2/(R_1 + R_2).