Ch 25: Resistors & DC CircuitsWorksheetSee all chapters
All Chapters
Ch 01: Units & Vectors
Ch 02: 1D Motion (Kinematics)
Ch 03: 2D Motion (Projectile Motion)
Ch 04: Intro to Forces (Dynamics)
Ch 05: Friction, Inclines, Systems
Ch 06: Centripetal Forces & Gravitation
Ch 07: Work & Energy
Ch 08: Conservation of Energy
Ch 09: Momentum & Impulse
Ch 10: Rotational Kinematics
Ch 11: Rotational Inertia & Energy
Ch 12: Torque & Rotational Dynamics
Ch 13: Rotational Equilibrium
Ch 14: Angular Momentum
Ch 15: Periodic Motion (NEW)
Ch 15: Periodic Motion (Oscillations)
Ch 16: Waves & Sound
Ch 17: Fluid Mechanics
Ch 18: Heat and Temperature
Ch 19: Kinetic Theory of Ideal Gasses
Ch 20: The First Law of Thermodynamics
Ch 21: The Second Law of Thermodynamics
Ch 22: Electric Force & Field; Gauss' Law
Ch 23: Electric Potential
Ch 24: Capacitors & Dielectrics
Ch 25: Resistors & DC Circuits
Ch 26: Magnetic Fields and Forces
Ch 27: Sources of Magnetic Field
Ch 28: Induction and Inductance
Ch 29: Alternating Current
Ch 30: Electromagnetic Waves
Ch 31: Geometric Optics
Ch 32: Wave Optics
Ch 34: Special Relativity
Ch 35: Particle-Wave Duality
Ch 36: Atomic Structure
Ch 37: Nuclear Physics
Ch 38: Quantum Mechanics

Concept #1: Intro to Kirchhoff's Loop Rule

Example #1: Direction of Current in Loop Equations

Example #2: Solving Circuits with Multiple Sources

Practice: For the circuit below, find the current through each of the 3 branches.

Concept #2: Combining Voltage Sources in Series

Example #3: Find Two Voltages (3 sources)

Practice: For the circuit below, calculate 

(a) the voltage V shown, and(b) the current through the 6-Ohm resistor.

Practice: For the circuit below, calculate the voltage across the 100-Ohm resistor.

Concept #3: How to Check Your Work (Kirchhoff's Rules)

Additional Problems
Present one junction and two loop rule equations that you could solve simultaneously to obtain values for the three currents, I1, I2 and I3, in the dc circuit shown below. Do not solve the equations!
Consider the circuit shown in the sketch. The current in the 8.0 Ω resistor is 2.0 A, in the direction shown. What is I1, the current through the 4.0 Ω resistor?  
Consider the circuit shown in the sketch. The current in the 8.0 Ω resistor is 2.0 A, in the direction shown. What is the resistance of the resistor R3?
In the following figure, the configuration of resistors is known as a Wheatstone Bridge, which is used to measure the resistance of an unknown resistor (let's say R1 in our case). This is accomplished by varying the resistance of one of the outside resistors (let's say R5 in our case) until the resistance through the center of the bridge is zero. For our bridge, the battery produces 20 V and the fixed resistances are R2 = 10 Ω, R3 = 20 Ω, and R4 = 30 Ω. If the current through the center of the bridges is zero, and the current produced by the battery is 1 A, when R5 reaches a value of 15 Ω, what is the unknown resistance?
In the following circuit, the resistances are R 1 = 5 Ω, R2 = 10 Ω, R3 = 10 Ω, R4 = 25 Ω, and R5 = 15 Ω. If the current through R3 is 1 A to the right, what is the voltage of the battery?
In the following circuit,V= 25 V, R 1 = 5 Ω, R2 = 15 Ω, R3 = 25 Ω, R4 = 10 Ω, and R5 = 20 Ω. Find each current in the circuit.
The circuit shown in the sketch consists of a battery with emf ε and internal resistance r connected to two resistors as shown. R1 = 3.0 Ω, R2 = 6.0 Ω and r = 0.50 Ω. The terminal voltage Vab of the battery is 14.0 V. At what rate is electrical energy dissipated in  R2?
Consider the circuit shown in the sketch. The current in the 8.0 Ω resistor is   I1 = 3.0 A. What is the current through 4.0 Ω resistor?
The circuit shown in the sketch consists of a battery with emf ε and internal resistance r connected to two resistors as shown. R1 = 3.0 Ω, R2 = 6.0 Ω and r = 0.50 Ω. The terminal voltage Vab of the battery is 14.0 V.  What is the emf of the battery?
Consider the circuit shown in the sketch. What is the current through the 3.0 Ω resistor?
Consider the circuit shown in the sketch. What is the current through the 10.0 V battery?
A 10 Ω lightblub is connected to a variable voltage source. The way to vary the voltage of the source is not by altering the source itself, but rather by altering the internal resistance of the source, thus changing the terminal voltage, Vab. If the internal resistance of the source were zero, the termal voltage would be 120 V, and at some value r, the power output by the lightbulb is 100 W. (a) What is the terminal voltage, Vab? (b) If the terminal voltage were doubled, what would be the power output by the lightbulb? (c) What is the internal resistance, r, of the lightbulb when the power output is 100 W?
In Figure 3, consider the circuit sketched. The two batteries have negligible internal resistance and emf's  ε1 = 28.0 V and ε2 = 42.0 V. The three resistors have resistance  R1 = 2.00 Ω, R2 = 5.00 Ω, R3 = 1.00 Ω. Calculate the potential difference Va - Vb between points a and b.
What current will a 3.0 V battery cause to flow if it is short circuited? The internal resistance of the battery is 14.0 Ω. A) 42 A B) 130 A C) 0.21 A D) 4.7 A
Your car will not start. You open the hood and measure the voltage on the battery as 14 volts. You then have a friend try to start the car while you continue to monitor the voltage on the battery. When your friend turns the ignition key, the battery voltage drops to 7.5 volts. If the car is drawing 150 A from the battery, what is the battery's internal resistance? a. 0.01 Ω b. 0.14 Ω c. 0.04 Ω d. 0.09 Ω e. 0.05 Ω
The circuit shown in the figure has resistors R1 = 10 Ω, R2 = 20 Ω, R3 = 30 Ω, ε1 = 10 V, ε2 = 20 V. What is the current through resistor R1? (1) 1 A (2) 2 A (3) 1.5 A (4) 0.5 A (5) 0.67 A
The two light bulbs in the circuit on the right are identical. When the switch is closed, A) they both go out B) the intensities of both A and B decrease C) the intensities of both A and B increase D) nothing changes E) some combination of (A)-(C) happens
The following equation is true for the right loop of the circuit. a) E2 - I3R3 - I2R2 - E1 = 0 b) -E2 - I3R3 - I2R2 - E1 = 0 c) E2 - I3R3 - I2R2 + E1 = 0 d) E2 + I3R3 + I2R2 - E1 = 0 e) None of the above
The following equation is true for the right loop if the circuit. E2 – I3R3 – I2R2 – E1 = 0 –E2 – I3R3 – I2R2 – E1 = 0 E2 – I3R3 – I2R2 + E1 = 0 E2 + I3R3 + I2R2 – E1 = 0 None of the above  
For the circuit below, determine all the currents. Assume ε1 = 10 V, ε2 = 15 V, R 1 = R 3 = 40 Ω, R 2 = 60 Ω, R 4 = 100 Ω.