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Current, Resistance, and DC Circuits

Ohm's law, Kirchhoff's rules, and power in circuits

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Current, Resistance, and DC Circuits

Electric Current

Definition: I=dQdtI = \frac{dQ}{dt}

Units: 1 ampere (A) = 1 coulomb/second

Current density: J⃗=nqv⃗d\vec{J} = nq\vec{v}_d

where nn is charge carrier density, qq is charge per carrier, v⃗d\vec{v}_d is drift velocity.

I=∫J⃗⋅dA⃗I = \int \vec{J} \cdot d\vec{A}

Resistance and Ohm's Law

Ohm's law: V=IRV = IR

Resistance: R=ρLAR = \frac{\rho L}{A}

where:

  • ρ\rho = resistivity (material property)
  • LL = length
  • AA = cross-sectional area

Conductivity: σ=1/ρ\sigma = 1/\rho

Microscopic Ohm's law: J⃗=σE⃗\vec{J} = \sigma\vec{E}

Temperature Dependence

ρ=ρ0[1+α(T−T0)]\rho = \rho_0[1 + \alpha(T - T_0)]

where α\alpha is temperature coefficient of resistivity.

Power

Power dissipated: P=IV=I2R=V2RP = IV = I^2R = \frac{V^2}{R}

Energy: E=PtE = Pt

Resistors in Series

Same current through each:

Req=R1+R2+⋯R_{eq} = R_1 + R_2 + \cdots

Vtotal=V1+V2+⋯V_{total} = V_1 + V_2 + \cdots

Resistors in Parallel

Same voltage across each:

1Req=1R1+1R2+⋯\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots

Two resistors: Req=R1R2R1+R2R_{eq} = \frac{R_1R_2}{R_1 + R_2}

Kirchhoff's Rules

Junction rule (current): ∑Iin=∑Iout\sum I_{in} = \sum I_{out}

(Charge conservation)

Loop rule (voltage): ∑ΔV=0\sum \Delta V = 0

(Energy conservation)

Sign Conventions

  • Traversing resistor with current: −IR-IR
  • Traversing resistor against current: +IR+IR
  • Traversing battery from −- to ++: +E+\mathcal{E}
  • Traversing battery from ++ to −-: −E-\mathcal{E}

EMF and Internal Resistance

Real battery has internal resistance rr:

Vterminal=E−IrV_{terminal} = \mathcal{E} - Ir

Maximum power to load: When Rload=rR_{load} = r (matched impedance)

Multiloop Circuits

Strategy:

  1. Assign current to each branch
  2. Apply junction rule
  3. Apply loop rule to independent loops
  4. Solve system of equations

Ammeter and Voltmeter

Ammeter: Measures current, low resistance (ideally 0), in series

Voltmeter: Measures voltage, high resistance (ideally ∞), in parallel

Wheatstone Bridge

Balanced when: R1R2=R3R4\frac{R_1}{R_2} = \frac{R_3}{R_4}

No current through galvanometer.

📚 Practice Problems

1Problem 1easy

❓ Question:

A cylindrical wire of radius r = 1.5 mm and length L = 2.0 m has resistivity ρ = 1.7 × 10⁻⁸ Ω·m. (a) Find the resistance of the wire. (b) If a current I = 5.0 A flows through it, find the current density J. (c) Find the electric field inside the wire.

💡 Show Solution

Given:

  • r = 1.5 mm = 1.5 × 10⁻³ m
  • L = 2.0 m
  • ρ = 1.7 × 10⁻⁸ Ω·m
  • I = 5.0 A

(a) Resistance:

Cross-sectional area: A=πr2=π(1.5×10−3)2=7.07×10−6 m2A = \pi r^2 = \pi (1.5 \times 10^{-3})^2 = 7.07 \times 10^{-6} \text{ m}^2

R=ρLA=(1.7×10−8)(2.0)7.07×10−6R = \frac{\rho L}{A} = \frac{(1.7 \times 10^{-8})(2.0)}{7.07 \times 10^{-6}}

R=4.81×10−3 Ω=4.81 mΩR = \boxed{4.81 \times 10^{-3} \text{ Ω} = 4.81 \text{ mΩ}}

(b) Current density:

J=IA=5.07.07×10−6J = \frac{I}{A} = \frac{5.0}{7.07 \times 10^{-6}}

J=7.07×105 A/m2J = \boxed{7.07 \times 10^5 \text{ A/m}^2}

(c) Electric field:

Using Ohm's law in differential form: J⃗=σE⃗\vec{J} = \sigma \vec{E} where σ=1/ρ\sigma = 1/\rho: E=ρJ=(1.7×10−8)(7.07×105)E = \rho J = (1.7 \times 10^{-8})(7.07 \times 10^5)

E=0.012 V/m=12 mV/mE = \boxed{0.012 \text{ V/m} = 12 \text{ mV/m}}

Alternatively: E=V/LE = V/L where V=IR=(5.0)(4.81×10−3)=0.024V = IR = (5.0)(4.81 \times 10^{-3}) = 0.024 V

2Problem 2medium

❓ Question:

In the circuit shown, a 24 V battery is connected to three resistors: R₁ = 6.0 Ω, R₂ = 4.0 Ω, and R₃ = 12 Ω. R₁ and R₂ are in parallel, and this combination is in series with R₃. Find: (a) the equivalent resistance, (b) the current through R₃, and (c) the power dissipated by each resistor.

💡 Show Solution

Given:

  • V = 24 V
  • R₁ = 6.0 Ω, R₂ = 4.0 Ω, R₃ = 12 Ω

(a) Equivalent resistance:

R₁ and R₂ in parallel: 1R12=1R1+1R2=16.0+14.0=512\frac{1}{R_{12}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{6.0} + \frac{1}{4.0} = \frac{5}{12}

R12=2.4 ΩR_{12} = 2.4 \text{ Ω}

Total resistance: Req=R12+R3=2.4+12=14.4 ΩR_{eq} = R_{12} + R_3 = 2.4 + 12 = \boxed{14.4 \text{ Ω}}

(b) Current through R₃:

Total current from battery: Itotal=VReq=2414.4I_{total} = \frac{V}{R_{eq}} = \frac{24}{14.4}

I3=Itotal=1.67 AI_3 = I_{total} = \boxed{1.67 \text{ A}}

(In series, all current flows through R₃)

(c) Power dissipated:

Voltage across parallel combination: V12=V−V3=V−I3R3=24−(1.67)(12)=4.0 VV_{12} = V - V_3 = V - I_3 R_3 = 24 - (1.67)(12) = 4.0 \text{ V}

P1=V122R1=(4.0)26.0=2.67 WP_1 = \frac{V_{12}^2}{R_1} = \frac{(4.0)^2}{6.0} = \boxed{2.67 \text{ W}}

P2=V122R2=(4.0)24.0=4.0 WP_2 = \frac{V_{12}^2}{R_2} = \frac{(4.0)^2}{4.0} = \boxed{4.0 \text{ W}}

P3=I32R3=(1.67)2(12)=33.3 WP_3 = I_3^2 R_3 = (1.67)^2(12) = \boxed{33.3 \text{ W}}

Total power: 2.67 + 4.0 + 33.3 = 40 W = VI ✓

3Problem 3hard

❓ Question:

Two batteries (ε₁ = 12 V with internal resistance r₁ = 0.5 Ω, and ε₂ = 6.0 V with r₂ = 0.3 Ω) are connected in parallel to an external load R = 3.0 Ω. Use Kirchhoff's laws to find: (a) the current through each battery, (b) the current through R, and (c) the terminal voltage across R.

💡 Show Solution

Given:

  • ε₁ = 12 V, r₁ = 0.5 Ω
  • ε₂ = 6.0 V, r₂ = 0.3 Ω
  • R = 3.0 Ω

Setup:

Let I₁ = current from battery 1 (positive right) Let I₂ = current from battery 2 (positive right) Let I = current through R (positive down)

Kirchhoff's current law at top junction: I1+I2=I...(1)I_1 + I_2 = I \quad \text{...(1)}

(a) & (b) Finding currents:

Loop through battery 1 and R (clockwise): ε1−I1r1−IR=0\varepsilon_1 - I_1 r_1 - IR = 0 12−0.5I1−3I=0...(2)12 - 0.5I_1 - 3I = 0 \quad \text{...(2)}

Loop through battery 2 and R (clockwise): ε2−I2r2−IR=0\varepsilon_2 - I_2 r_2 - IR = 0 6−0.3I2−3I=0...(3)6 - 0.3I_2 - 3I = 0 \quad \text{...(3)}

From (2): I=12−0.5I13I = \frac{12 - 0.5I_1}{3}

From (3): I=6−0.3I23I = \frac{6 - 0.3I_2}{3}

Setting equal: 12−0.5I1=6−0.3I212 - 0.5I_1 = 6 - 0.3I_2 6=0.5I1−0.3I2...(4)6 = 0.5I_1 - 0.3I_2 \quad \text{...(4)}

From (1): I2=I−I1I_2 = I - I_1

Substitute into (4): 6=0.5I1−0.3(I−I1)=0.5I1−0.3I+0.3I1=0.8I1−0.3I6 = 0.5I_1 - 0.3(I - I_1) = 0.5I_1 - 0.3I + 0.3I_1 = 0.8I_1 - 0.3I

From (2): I=4−0.167I1I = 4 - 0.167I_1

Substitute: 6=0.8I1−0.3(4−0.167I1)6 = 0.8I_1 - 0.3(4 - 0.167I_1) 6=0.8I1−1.2+0.05I16 = 0.8I_1 - 1.2 + 0.05I_1 7.2=0.85I17.2 = 0.85I_1

I1=8.47 AI_1 = \boxed{8.47 \text{ A}}

I=4−0.167(8.47)=2.58 AI = 4 - 0.167(8.47) = \boxed{2.58 \text{ A}}

I2=2.58−8.47=−5.89 AI_2 = 2.58 - 8.47 = \boxed{-5.89 \text{ A}}

(Negative means battery 2 is being charged!)

(c) Terminal voltage:

V=IR=(2.58)(3.0)=7.74 VV = IR = (2.58)(3.0) = \boxed{7.74 \text{ V}}

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Ohm's law, Kirchhoff's rules, and power in circuits
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