Ohm’s & Kirchoff’s Laws, Resistive Loads and Meters

BME253L - Fall 2025

Dr. Mark Palmeri, M.D., Ph.D.

Duke University

August 27, 2025

Learning Objectives

• Analyze Power and Energy Flow in Circuits

• Define Ohm’s Law

• Define Kirchoff’s Voltage and Current Laws

• Resistive Loads

• Voltage and Current Measurement Devices

Review from Last Lecture

\[ i = \frac{dq}{dt} \left[\frac{C}{s}\right] = [A] (Ampere) \]

\[ V = \frac{dW}{dq} \left[\frac{J}{C}\right] = [V] (Volt) \]

\[ P = \frac{dw}{dt} = \frac{dW}{dq} \frac{dq}{dt} = VI \left[W = \frac{J}{s}\right] \]

• Power is the capacity to do work per unit time, where:

\[ Work = Q \times V \]

which is the transfer of energy associated with moving charge across a potential difference.

• Sources provide energy to a circuit (\(P < 0\)).

• Loads consume energy from a circuit (\(P > 0\)).

• Postive power is when postive current flows from \((+) -> (-)\) voltage drop across a circuit element:

Power Source/Load Example

Given:

\[ \begin{gather} P_1 = -100 W \\ P_2 = 50 W \\ P_3 = 30 W \\ v_1 = 10 V \\ v_2 = 5 V \\ i_3 = 6 A \end{gather} \]

We don’t know what these circuit components are, but given power/voltage/current associated with them, we can determine if they are sources or loads, and solve for the unknown quantities.

Circuit Element 1

• Source or load?
• Solve for \(i_1\): \[ \begin{gather} P_1 = v_1 i_1 \\ -100 W = 10 V (i_1) \\ i_1 = -10 A \end{gather} \]
• Element 1 is a source!

Circuit Element 2

• Source or load?
• Solve for \(i_2\): \[ \begin{gather} P_2 = v_w i_2 \\ 50 W = 5 V (i_2) \\ i_2 = 10 A \end{gather} \]
• Element 2 is a load!

Kirchoff’s Current Law (KCL)

• Conservation of energy means that for a given node, all current flowing in and out of the node must be conserved.

\[ \begin{gather} \Sigma i_{in} = \Sigma i_{out} \end{gather} \]

• For Node A, what currents are “in” and “out”?

• As drawn (arbitrary), both are “out”, so:

\[ \begin{gather} \Sigma i_{in} = 0 \\ \Sigma i_{out} = i_1 + i_2 = -10 A + 10 A = 0 ✅ \end{gather} \]

• In this circuit, \(i_1 = -i_2\), which makes sense because they are the same magnitude, but flowing in different directions relative.

Kirchoff’s Current Law (KCL)

• Generalized form of current conservation at a node:

\[ \Sigma_{n=1}^N i_n = 0 \]

• You have to be very careful to correctly capture current direction (“in”/“out”) and magnitude.

Circuit Element 3

Source or load?

\[ v_3 = \frac{P_3}{i_3} = \frac{30 W}{6 A} = 5 V \]

Positive current flow “down” a voltage drop, so load!

Circuit Element 4

Source or load? You tell me… do we know enough?

Yes, we do!

• Conservation of energy!

• \(\Sigma_n P_n = 0\)

• \(P_1 + P_2 + P_3 + P_4 = 0\)

• \(P_4\) = 20 W (positive, so a load!)

What are \(v_4\) and \(i_4\)?

• Assume that wires in circuits are ideal conductors (i.e., no energy loss).

• \(v_B\) = \(v_C\) (same node)

• Apply KCL:

\[ \begin{gather} i_{in} = i_{out} \\ i_2 + i_4 = i_3 \\ i_4 = i_3 - i_2 = 6 A - 10 A = -4 A \\ v_4 = \frac{P_4}{i_4} = \frac{20 W}{-4 A} = -5 V \end{gather} \]

Did we have to solve for \(i_4\)?

• No!

• \(v_B = v_C\) (ideal conductor assumption; same node)

• Circuit elements 3 & 4 share common node \(D\).

• As drawn on the circuit:

\[ \begin{gather} v_3 = v_B - v_D \\ v_4 = v_D - v_C = v_D - v_B = -v_3 = -5 V \end{gather} \]

• Then you can solve for \(i_4 = \frac{P_4}{v_4} = \frac{20 W}{-5 V} = -4 A\).

Kirchoff’s Voltage Law (KVL)

• The last example demonstrated another fundamental circuit analysis law:

\[ \Sigma_{n=1}^{N} v_n = 0 \]

• For a closed loop of a circuit (continuous path of conduction), the sum of all voltage gains = sum of all voltage losses.

\[ \Sigma v_{source} = \Sigma v_{load} \]

Does our circuit obey KVL?

• We have 3 (not 2) closed loops:

  • \(D \rightarrow A \rightarrow (B + C) \rightarrow D\)

  \[ \begin{gather} \Sigma v = 0 = -v_1 + v_2 - v_4\\ 0 = -10 V + 5 V - (-5 V) = 0 ✅ \end{gather} \]

  • \(D \rightarrow A \rightarrow B \rightarrow D\) \[ \begin{gather} \Sigma v = 0 = -v_1 + v_2 + v_3\\ 0 = -10 V + 5 V + 5 V = 0 ✅ \end{gather} \]

  • \(D \rightarrow B \rightarrow C \rightarrow D\)

  \[ \begin{gather} \Sigma v = 0 = -v_3 - v_4 \\ 0 = -5 V - (-5 V) = 0 ✅ \end{gather} \]

What happens if we reverse the loops?

• You tell me.

• Prove it.

Summary of Connection Constraints

Important

• Balance of Power: \(\Sigma_n P_n = 0\)

• Conservation of Current at a Node (KCL): \(\Sigma_n i_n = 0\)

• Conservation of Voltage around a Closed Loop (KVL): \(\Sigma_n v_n = 0\)

What do these laws tell us…

Parallel Devices

Devices in parallel have the same voltage difference across them:

\[ v_1 = v_2 \]

Serial Devices

Devices in series have the same current flowing through them.

\[ i_1 = i_2 \]

Resistive Loads

• Resistors are dissipative loads (typically as heat).

• Conductive wire before/after the resistor element is considered “ideal” (non-dissipative).

• \(\rho_{carbon} = 3.5 \times 10^{-5} \Omega \cdot m\)

Ohm’s Law

\[ \begin{gather} R = \frac{V}{I}\\ V = IR \end{gather} \]

• Applicable over the linear range of “\(I-V\)” behavior (i.e., resistance, \(R\) doesn’t change as a a function of voltage or current).

• \([\Omega]\) Ohm \(\rightarrow 1 \Omega = 1 \frac{V}{A}\)

• Resistors always dissipate power (\(P > 0\)):

\[ \begin{gather} P_R = i_R v_R = i_R (i_R R) = i_R^2 R \geq 0\\ P_R = i_R v_R = \left(\frac{v_R}{R}\right)v_R = \frac{v_R^2}{R} \geq 0 \end{gather} \]

Conductance

• Conductance (\(G\)) is the reciprocal of resistance.

• Unit: Siemen [\(S\)]

• \(1 S = 1 \frac{A}{V}\)

• \(i = G v\)

Conductivity of Materials

Resistive Power

• \(P_R = i^2R = \frac{v^2}{R}\)

• Quadratic dependence on \(i\) and \(v\).

• Resistors have a maximum power rating that they can tolerate before they are damaged.m

Ceramic Resistor Labeling

Switches

Switches are mechanical devices that interrupt or divert current flow in a circuit.

• Open interrupter switches create an open circuit:

  • \(R \rightarrow \infty\)

  • \(i = 0\) (regardless of \(v\), to a point)

• Closed interrupter switches create a short circuit:

  • \(R = 0\)

  • \(v = 0\) (regardless of \(i\))

• \(P_{open} = 0 = P_{short}\) (why?)

Ideal Sources

• For an ideal voltage source, the source current is a function of the attached circuit resistance.

• For an ideal current course, the source voltage is a function of the attached circuit resistance.

“Real” Source

• Real sources have internal resistance.

• Ideally:

  • Internal resistance of a voltage source \(\rightarrow 0\).

  • Internal resistance of a current source \(\rightarrow \infty\).

• Why?

Tip

Think KCL and KVL…

Implications of Internal Source Resistance

• “Steals” power from being delivered to attached load.

• Generates parasitic heat internally.

“Bad” Loads

Shorted voltage source?

What happens if the load attached to a voltage source is a short circuit (i.e., \(R_L = 0\))?

Open current source?

What happens if the load attached to a current source is an open circuit (i.e., \(R_L \rightarrow \infty\))?

Measuring Voltage, Current & Resistance

Summary of Circuit Analysis

• Circuits are a connection collection of circuit elements that deliver, dissipate and store energy.

• Ohm’s Law, KCL & KVL can be used to solve for the \(v\) and \(i\) quantities for circuit components.

• Define directions (arbitrary) for current flow for all circuit components.

Tip

You do not have to try to “figure out” if a circuit component is a source or load; just assign the current direction and stick with it. The sign of the current will tell you what you need for its given direction.

• Label the voltage polarity of all sources and passive elements.

Tip

Have your currents point into the (+) voltage terminal of circuit components, and then if \(i > 0\), you know that \(P > 0\).

• Compute circuit components powers.

  • \((+) i\) flows into \((+) v \rightarrow P > 0\)

  • \((+) i\) flows out of \((+) v \rightarrow P < 0\)

• \(\Sigma_n P_n = 0\) ✅