Worksheet 16D - Harder Series-Parallel calculations - Solutions and Commentary

Introduction

The voltage drop when a power supply (in this case a battery) is placed under load can be modelled as the effect of an internal resistance.

The power supply has an internal EMF, which is the "no-load" (\( I_{\mathrm{L}} = 0 \)) voltage. Load current will cause a voltage drop across the internal resistance, lowering the voltage seen at the supply terminals.

The effect of the internal resistance may be modelled using voltage division, using the internal resistance, and the total resistance.

Calculations using Voltage Division

The general formula for supply voltage with internal resistance is as follows:

\( V_{\mathrm{s}} = \mathcal{E} \cdot \frac{R_{\mathrm{L}}}{R_{\mathrm{L}} + R_{\mathrm{Int}}} \)

where \( V_{\mathrm{s}} \) is the supply voltage seen at the supply terminals, \(\mathcal{E} \) is the internal EMF of the power supply, \( R_{\mathrm{L}} \) is the total load resistance, and \( R_{\mathrm{Int}} \) is the internal resistance of the power supply.

It is not possible to measure \( R_{\mathrm{Int}} \) directly, as it is not possible to separate it from the EMF generating mechanism (e.g. in a battery the electrolyte may contribute a resistive effect at the same time as it assists generating the EMF).

The theoretical method of measuring \( R_{\mathrm{Int}} \) is given by the formula below.

\( R_{\mathrm{Int}} = \frac{V_{\mathrm{oc}}}{I_{\mathrm{sc}}} \)

where \( R_{\mathrm{Int}} \) is the internal resistance, \( V_{\mathrm{oc}} \) is the open-curcuit voltage (no-load voltage), and \( I_{\mathrm{sc}} \) is the short-circuit current. This method is generally not used on real sources, since the short circuit current is often dangerously high. Further, many power supplies (especially current-limited power supplies) have a value of \( R_{\mathrm{Int}} \) that depends on the load current. Most methods of measuring \( R_{\mathrm{Int}} \) rely on measuring the voltage drop for a known load current.

The internal voltage drop is given in the equation below.

\( V_{\mathrm{Int}} = \mathcal{E} - V_{\mathrm{s}} \)

where \( V_{\mathrm{Int}} \) is the internal voltage drop, and \( \mathcal{E} \) and \( V_{\mathrm{s}} \) have the same meanings as above.

Calculations using Load Current

If the load current is known, the calculations can be simplified.

The general formula for supply voltage with internal resistance is as follows:

\( V_{\mathrm{s}} = \mathcal{E} - I_{\mathrm{L}} R_{\mathrm{Int}} \)

where \( V_{\mathrm{s}} \) is the supply voltage seen at the supply terminals, \(\mathcal{E} \) is the internal EMF of the power supply, \( I_{\mathrm{L}} \) is the load current, and \( R_{\mathrm{Int}} \) is the internal resistance of the power supply.

It is not possible to measure \( R_{\mathrm{Int}} \) directly, as it is not possible to separate it from the EMF generating mechanism (e.g. in a battery the electrolyte may contribute a resistive effect at the same time as it assists generating the EMF).

The theoretical method of measuring \( R_{\mathrm{Int}} \) is given by the formula below.

\( R_{\mathrm{Int}} = \frac{\mathcal{E} - V_{\mathrm{s}}}{I_{\mathrm{L}}} \)

where \( R_{\mathrm{Int}} \) is the internal resistance, and all other symbols have the same meanings as above.

\( R_{\mathrm{Int}} = I_{\mathrm{L}} R_{\mathrm{Int}} \)

where \( V_{\mathrm{Int}} \) is the internal voltage drop, and all other symbols have the same meanings as above.

Mixing and Matching Calculations

It is possible to mix and match calculations to get voltages, currents and resistances. As long as the method complies with Ohm's Law and Kirchhoff's Laws, it will give the correct results.

Question 1

A marked-up version of the circuit is shown below.

The objective is to find the following quantities:

• The voltages \( V_{R1} \), \( V_{R2} \), \( V_{R3} \), \( V_{\mathrm{Int}} \) and \( V_{\mathrm{s}} \).
  

Solution - Load Current

This method uses the load current to calculate all the voltage drops across the resistors.

\( I_{\mathrm{L}} = \frac{\mathcal{E}}{R_{\mathrm{L}} + R_{\mathrm{Int}}} = \frac{65}{(15 + 20 + 25) + 5} = \frac{65}{65} = \) 1 A.

The load current can now be used to calculate all the resistor voltage drops.

• \( V_{R1} = I_{\mathrm{L}} R_1 = 1 \cdot 15 = \) 15 V #.
  
  
• \( V_{R2} = I_{\mathrm{L}} R_2 = 1 \cdot 20 = \) 20 V #.
  
  
• \( V_{R3} = I_{\mathrm{L}} R_3 = 1 \cdot 25 = \) 25 V #.
  
  
• \( V_{\mathrm{Int}} = I_{\mathrm{L}} R_{\mathrm{Int}} = 1 \cdot 5 = \) 5 V #.
  
  
• \( V_{\mathrm{s}} = \mathcal{E} - V_{\mathrm{Int}} = 65 - 5 = \) 60 V #.

Solution - Voltage Division

The general formula for the voltage drop across any resistor is \( V_R = \mathcal{E} \cdot \frac{R}{R_{\mathrm{L}} + R_{\mathrm{Int}}} \).

The quantities are as follows:

• \( \mathcal{E} \) = 65 V.
  
  
• \( R_{\mathrm{L}} = 15 + 20 + 25 = \) 60 Ω.
  
  
• \( R_{\mathrm{L}} + R_{\mathrm{Int}} = (15 + 20 + 25) + 5 = \) 65 Ω.
  

The calculations themselves are as follows.

• \( V_{R1} = 65 \cdot \frac{15}{65} = \) 15 V #.
  
  
• \( V_{R2} = 65 \cdot \frac{20}{65} = \) 20 V #.
  
  
• \( V_{R3} = 65 \cdot \frac{25}{65} = \) 25 V #.
  
  
• \( V_{\mathrm{Int}} = 65 \cdot \frac{5}{65} = \) 5 V #.
  
  
• \( V_{\mathrm{s}} = 65 \cdot \frac{60}{65} = \) 60 V #.

Question 2

Since all voltages are measured with respect to 'A', point 'A' will be considered to be at zero volts. A voltage \( V_{\mathrm{A-x}} \) is a voltage measured with a multimeter with its positive lead at point 'x', and its negative lead at point 'A'.

The circuit is shown below. In order to reduce visual clutter, the connections of the meter are not shown. In order to ease the calculations, a new node 'D' has been added.

NOTE: The circuit is drawn in a somewhat confusing manner. The wires crossing at an angle are not connected. Dots have been added to clarify which wires are actually connected together.

The circuit is shown re-drawn with the cross-over eliminated.

This version of the circuit is substantially easier to analyse.

Solution - Method 1 - Back-Substitution

The back-substitution method is purely based on calculating nodal voltages. No branch currents are calculated at all (at least not directly). All voltages are calculated by voltage division and back-substitution.

This method is based on back-substitution. The nodal voltages are related to each other, then the voltages are solved using back-substitution from known voltages. It may be thought of as a "chain of calculations", where one quantity is solved, this is used to solve the next quantity and so on.

The voltage at the nodes B, C and D can be defined as follows using voltage division:

• \( V_{\mathrm{A-C}} = V_{\mathrm{A-D}} \cdot \frac{50}{50 + 50} = \frac{1}{2} V_{\mathrm{A-D}} \).
  
  
• \( V_{\mathrm{A-D}} = V_{\mathrm{A-B}} \cdot \frac{100 \parallel (50 + 50)}{25 + 25 + 100 \parallel (50 + 50)} = V_{\mathrm{A-B}} \cdot \frac{50}{50 + 50} = \frac{1}{2} V_{\mathrm{A-B}} \).
  
  
• \( V_{\mathrm{A-B}} = \mathcal{E} \cdot \frac{R_{\mathrm{T}}}{R_{\mathrm{Int}} + R_{\mathrm{T}}} = 102 \cdot \frac{R_{\mathrm{T}}}{2 + R_{\mathrm{T}}} \).
  
  

In order to go further, the total load resistance \( R_{\mathrm{T}} \) must be calculated.

\( R_{\mathrm{T}} = 25 + 25 + (50 + 50) \parallel 100 = 50 + 100 \parallel 100 = 50 + 50 = \) 100 Ω.

We can now calculate the voltages \( V_{\mathrm{A-B}} \), \( V_{\mathrm{A-D}} \), and \( V_{\mathrm{A-C}} \) by back-substitution.

• \( V_{\mathrm{A-B}} = 102 \cdot \frac{100}{2 + 100} = 102 \cdot \frac{100}{102} = \) 100 V #.
  
  
• \( V_{\mathrm{A-D}} = \frac{1}{2} V_{\mathrm{A-B}} = \frac{1}{2} \cdot 100 = \) 50 V.
  
  
• \( V_{\mathrm{A-C}} = \frac{1}{2} V_{\mathrm{A-D}} = \frac{1}{2} \cdot 50 = \) 25 V #.

Solution - Method 2 - Kirchhoff's Current Law and Kirchhoff's Voltage Law

This method is based on calculating currents, then splitting these currents through the circuit.

In order to do this method, the total load resistance as seen by the internal EMF \( \mathcal{E} \) must be known first.

The load resistance \( R_{\mathrm{T}} \) is calculated as follows.

\( R_{\mathrm{T}} = 25 + 25 + (50 + 50) \parallel 100 = 50 + 100 \parallel 100 = 50 + 50 = \) 100 Ω.

The load resistance can be used to calculate the current flow, based on the total load and internal resistances as seen by internal EMF \( \mathcal{E} \).

\( I_{\mathrm{L}} = \frac{\mathcal{E}}{R_{\mathrm{Int}} + R_{\mathrm{T}}} = \frac{102}{2 + 100} = \frac{102}{102} = \) 1 A.

The voltage between A and B can be calculated using Kirchoff's Voltage Law.

\( V_{\mathrm{A-B}} = \mathcal{E} - I_{\mathrm{L}} R_{\mathrm{Int}} = 102 - 1 \cdot 2 = 102 - 2 = \) 100 V #.

The voltage between A and D can be calculated using Kirchoff's Voltage Law.

\( V_{\mathrm{A-D}} = V_{\mathrm{A-B}} - I_{\mathrm{L}} \cdot (25 + 25 ) = 100 - 1 \cdot 50 = 100 - 50 = \) 50 V.

Looking at node D, the current splits into two branches. The branch current flowing through the two 50 Ω resistors is calculated below.

\( I = \frac{V_{\mathrm{A-D}}}{50 + 50} = \frac{50}{100} = \) 0.5 A.

The voltage at node \( V_{\mathrm{A-C}} \) can be calculated directly using Ohm's Law.

\( V_{\mathrm{A-C}} = I \cdot 50 = 50 \cdot 0.5 = \) 25 V #.

Question 3

The circuit is shown below, with resistances that will be used in the solution. The resistances shown are the resistances "down" to node A from the node shown. They do not include the "upstream" resistances e.g. \( R_4 \) does not include the 5 Ω resistor.

This is a complex ladder circuit, but the voltages can be determined by voltage division back-substitution, or calculating branch currents and subtracting voltage drops.

This circuit is right on the limit of what you would work out by hand. Any more complex circuits should generally be solved using computer analysis. The primary reasons is that maintaining the precision through intermediate calculations becomes difficult, and it is easy to make a mistake anywhere during the calculations that makes all subsequent calculations incorrect.

Some nodal resistances are pre-calculated below. The resistances at a particular node are measured "directly" from the node to ground through the set of resistances with no branches. For example, \( R_{\mathrm{C}} \) is calculated from the set of the three 10 Ω, 15 Ω and 10 Ω resistors to node A.

• \( R_{\mathrm{C}} = 10 + 15 + 10 = \) 40 Ω.
• \( R_{\mathrm{D}} = 10 + 15 = \) 25 Ω.
• \( R_{\mathrm{E}} = 15 \parallel 15 = \) 7.5 Ω.
• \( R_{\mathrm{F}} = \) 15 Ω.

Solution - Method 1 - Back-Substitution

The back-substitution method is purely based on calculating nodal voltages. No branch currents are calculated at all (at least not directly). All voltages are calculated by voltage division and back-substitution.

The circuit is continually "collapased", until all the resistances in the circuit are lumped into a single resistance.

The first step is shown below. Every purely series-parallel resistor network that is directly connected to node A has been simplified.

We also calculate some resistances, which will help us later on.

• \( R_{\mathrm{C}} = 10 + 15 + 10 = \) 40 Ω.
• \( R_{\mathrm{D}} = 10 + 15 = \) 25 Ω.
• \( R_{\mathrm{E}} = 15 \parallel 15 = \) 7.5 Ω.
• \( R_{\mathrm{F}} = \) 15 Ω.
• \( R_1 = 10 + R_{\mathrm{E}} = 10 + 15 \parallel 15 = \) 17.5 Ω.
  
  

Nodes E and F have been eliminated, but we can use voltage division to "recover" the voltages at the eliminated nodes. The equations are shown below.

• \( V_{\mathrm{A-F}} = V_{\mathrm{A-C}} \cdot \frac{R_{\mathrm{F}}}{R_{\mathrm{C}}}= V_{\mathrm{A-C}} \cdot \frac{15}{40} = \frac{3}{8} V_{\mathrm{A-C}} \).
  
  
• \( V_{\mathrm{A-E}} = V_{\mathrm{A-D}} \cdot \frac{R_{\mathrm{E}}}{R_1} = V_{\mathrm{A-B}} \cdot \frac{7.5}{17.5} = \frac{3}{7} V_{\mathrm{A-D}} \).

The process is repeated again, producing the circuit below.

We also calculate some resistances, which will help us later on.

• \( R_{\mathrm{C}} = \) 40 Ω.
  
  
• \( R_{\mathrm{D}} = \) 25 Ω.
  
  
• \( R_{\mathrm{E}} = \) 7.5 Ω.
  
  
• \( R_{\mathrm{F}} = \) 15 Ω.
  
  
• \( R_1 = \) 17.5 Ω.
  
  
• \( R_2 = R_1 \parallel R_{\mathrm{D}} = 25 \parallel 17.5 = \) 10.2941 Ω (6 sf).
  
  
• \( R_3 = 15 + R_2 = \) 25.2941 Ω (6 sf).
  
  

Nodes E and F have been eliminated, but we can use voltage division to "recover" the voltages at the eliminated nodes. The equations are shown below.

• \( V_{\mathrm{A-F}} = \frac{3}{8} V_{\mathrm{A-C}} \).
  
  
• \( V_{\mathrm{A-E}} = \frac{3}{7} V_{\mathrm{A-D}} \).
  
  
• \( V_{\mathrm{A-D}} = V_{\mathrm{A-C}} \cdot \frac{R_2}{R_3} = V_{\mathrm{A-B}} \cdot \frac{10.2941}{25.2941} = 0.406976 V_{\mathrm{A-C}} \).
  
  

The process is repeated again, producing the circuit below.

We also calculate some resistances, which will help us later on.

• \( R_{\mathrm{C}} = \) 40 Ω.
  
  
• \( R_{\mathrm{D}} = \) 25 Ω.
  
  
• \( R_{\mathrm{E}} = \) 7.5 Ω.
  
  
• \( R_{\mathrm{F}} = \) 15 Ω.
  
  
• \( R_1 = \) 17.5 Ω.
  
  
• \( R_2 = \) 10.2941 Ω.
  
  
• \( R_3 = \) 25.2941 Ω.
  
  
• \( R_4 = R_3 \parallel R_{\mathrm{C}} = 25.2941 \parallel 40 = \) 15.4955 Ω (6 sf).
  
  
• \( R_{\mathrm{L}} = R_4 + 5 = \) 20.4955 Ω (6 sf).
  
  

Nodes E and F have been eliminated, but we can use voltage division to "recover" the voltages at the eliminated nodes. The equations are shown below.

• \( V_{\mathrm{A-F}} = \frac{3}{8} V_{\mathrm{A-C}} \).
  
  
• \( V_{\mathrm{A-E}} = \frac{3}{7} V_{\mathrm{A-D}} \).
  
  
• \( V_{\mathrm{A-D}} = 0.406976 V_{\mathrm{A-C}} \).
  
  
• \( V_{\mathrm{A-C}} = V_{\mathrm{A-B}} \cdot \frac{R_4}{R_{\mathrm{L}}} = V_{\mathrm{A-B}} \cdot \frac{15.4955}{20.4955} = 0.803886 V_{\mathrm{A-B}} \).
  
  

The process is repeated again, producing the circuit below.

We also calculate some resistances, which will help us later on.

• \( R_{\mathrm{C}} = \) 40 Ω.
  
  
• \( R_{\mathrm{D}} = \) 25 Ω.
  
  
• \( R_{\mathrm{E}} = \) 7.5 Ω.
  
  
• \( R_{\mathrm{F}} = \) 15 Ω.
  
  
• \( R_1 = \) 17.5 Ω.
  
  
• \( R_2 = \) 10.2941 Ω.
  
  
• \( R_3 = \) 25.2941 Ω.
  
  
• \( R_4 = \) 15.4955 Ω (6 sf).
  
  
• \( R_{\mathrm{L}} = \) 20.4955 Ω.
  
  
• \( R_{\mathrm{Int}} = \) 5 Ω.
  
  

Nodes E and F have been eliminated, but we can use voltage division to "recover" the voltages at the eliminated nodes. The equations are shown below.

• \( V_{\mathrm{A-F}} = \frac{3}{8} V_{\mathrm{A-C}} \).
  
  
• \( V_{\mathrm{A-E}} = \frac{3}{7} V_{\mathrm{A-D}} \).
  
  
• \( V_{\mathrm{A-D}} = 0.406976 V_{\mathrm{A-C}} \).
  
  
• \( V_{\mathrm{A-C}} = V_{\mathrm{A-B}} \cdot \frac{R_4}{R_{\mathrm{L}}} = V_{\mathrm{A-B}} \cdot \frac{15.4955}{20.4955} = 0.803886 V_{\mathrm{A-B}} \).
  
  
• \( V_{\mathrm{A-B}} = \mathcal{E} \cdot \frac{R_{\mathrm{L}}}{R_{\mathrm{L}} + R_{\mathrm{Int}}} = V_{\mathrm{A-B}} \cdot \frac{20.4955}{25.4955} = 0.803886 \mathcal{E} \).
  
  

The voltage division equations are developed as follows, based on the series and parallel resistances present at each of the nodes B, C, D, E, and F with regard to the "upstream" node.

• \( V_{\mathrm{A-D}} = V_{\mathrm{A-C}} \cdot \frac{R_1 \parallel (10 + 15)}{R_1 \parallel (10 + 15) + 15} = V_{\mathrm{A-C}} \cdot \frac{10.2941}{10.2941 + 15} = V_{\mathrm{A-C}} \cdot \frac{10.2941}{25.2941} = 0.406976 V_{\mathrm{A-C}} \). We will also make note of the resistance \( R_2 = \) 25.2941 Ω for use in subsequent calculations.
  
  
• \( V_{\mathrm{A-C}} = V_{\mathrm{A-B}} \cdot \frac{R_{\mathrm{C-A}}}{R_{\mathrm{C-A}} + 5} = V_{\mathrm{A-C}} \cdot \frac{R_2 \parallel (15 + 10 + 15)}{R_2 \parallel (15 + 10 + 15) + 5} = V_{\mathrm{A-C}} \cdot \frac{15.4954}{15.4954 + 5} = V_{\mathrm{A-C}} \cdot \frac{15.4954}{20.4954} = 0.756044 V_{\mathrm{A-B}} \). We will also make note of the resistance \( R_3 = \) 20.4954 Ω for use in subsequent calculations.
  
  
• \( V_{\mathrm{A-B}} = \mathcal{E} \cdot \frac{R_3}{R_3 + R_{\mathrm{Int}}} = 147 \cdot \frac{20.4954}{20.4954 + 5} = 147 \cdot \frac{20.4954}{25.4954} = 147 \cdot 0.803886 = \) 118.171 V (6 sf) or 118.2 V (4 sf).
  
  

We can now calculate the voltages \( V_{\mathrm{A-B}} \) and \( V_{\mathrm{A-C}} \) by back-substitution.

• \( V_{\mathrm{A-B}} = \) 118.171 V (6 sf) or 118.2 V (4 sf).
  
  
• \( V_{\mathrm{A-C}} = 0.756044 V_{\mathrm{A-B}} = 0.756044 \cdot 118.171 = \) 89.3425 V (6 sf) or 89.34 V (4 sf) #.
  
  
• \( V_{\mathrm{A-F}} = \frac{3}{8} V_{\mathrm{A-C}} = \frac{3}{8} \cdot 89.3425 = \) 33.5034 V (6 sf) or 33.50 V (4 sf).
  
  
• \( V_{\mathrm{A-D}} = 0.406976 V_{\mathrm{A-C}} = 0.406976 \cdot 89.3425 = \) 36.3603 V (6 sf) or 36.36 V (4 sf).
  
  
• \( V_{\mathrm{A-E}} = \frac{3}{7} V_{\mathrm{A-D}} = \frac{3}{7} \cdot 36.3603 = \) 15.58 V (4 sf).

Solution - Method 2 - Kirchhoff's Current Law and Kirchhoff's Voltage Law

This method is based on calculating currents, then splitting these currents through the circuit.

In order to do this method, the total load resistance as seen by the internal EMF \( \mathcal{E} \) must be known first.

The load resistance \( R_{\mathrm{L}} \) is calculated as follows. The full calculation is shown below, but it will be broken up according to the resistances as shown above.

\( R_{\mathrm{L}} = 5 + (15 + 10 + 15) \parallel (15 + (10 + 15) \parallel (10 + 15 \parallel 15)) \)

This calculation is complicated to say the least, so it will be broken up into sub-calculations.

• \( R_{\mathrm{L}} = 5 + R_4 \)
• \( R_4 = R_{\mathrm{C}} \parallel R_3 = 40 \parallel R_3 \)
• \( R_3 = 15 + R_2 \)
• \( R_2 = R_{\mathrm{D}} \parallel R_1 = 25 \parallel R_1 \)
• \( R_1 = 10 + R_{\mathrm{E}} = 10 + 7.5 = \) 17.5 Ω

The resistance \( R_{\mathrm{L}} \) can be calculated by back-substitution.

• \( R_1 = \) 17.5 Ω
• \( R_2 = 25 \parallel 17.5 = \) 10.2941 Ω (6 sf).
• \( R_3 = 15 + R_2 = 15 + 10.2941 = \) 25.2941 Ω (6 sf).
• \( R_4 = 40 \parallel R_3 = 40 \parallel 25.2941 = \) 15.4955 Ω (6 sf).
• \( R_{\mathrm{L}} = 5 + 15.4955 = \) 20.4955 Ω (6 sf).

The load resistance can be used to calculate the current flow, based on the total load and internal resistances as seen by internal EMF \( \mathcal{E} \).

\( I_{\mathrm{L}} = \frac{\mathcal{E}}{R_{\mathrm{Int}} + R_{\mathrm{T}}} = \frac{147}{5 + 20.4955} = \frac{147}{25.4955} = \) 5.76572 A (6 sf).

The voltage between A and B can be calculated using Kirchoff's Voltage Law.

\( V_{\mathrm{A-B}} = \mathcal{E} - I_{\mathrm{L}} R_{\mathrm{Int}} = 147 - 5.76572 \cdot 5 = 147 - 28.8286 = \) 118.171 V (6 sf) or 118.2 (4 sf) #.

The voltage between A and C can be calculated using Kirchoff's Voltage Law.

\( V_{\mathrm{A-C}} = V_{\mathrm{A-B}} - I_{\mathrm{L}} \cdot 5 = 118.171 - 5.76572 \cdot 5 = 118.171 - 28.8286 = \) 89.3424 V (6 sf) or 89.34 V (4 sf) #.

Looking at node C, the current splits into two branches. The branch current flowing through the 15+10+15 Ω resistor branch (\( R_{\mathrm{C}} \)) is calculated below.

\( I = \frac{V_{\mathrm{A-C}}}{R_{\mathrm{C}}} = \frac{89.3424}{40} = \) 2.23356 A (6 sf).

Knowing this current gives an opportunity to calculate \( V_{\mathrm{A-F}} \).

\( V_{\mathrm{A-F}} = 15 \cdot I = 15 \cdot 2.23356 = \) 33.50 V (4 sf) #.

The branch current flowing between nodes C and D is calculated using Kirchhoff's Current Law.

\( I_{\mathrm{C-D}} = 5.76572 - 2.23356 = \) 3.53216 A (6 sf).

This current is used to calculate the voltage \( V_{\mathrm{A-D}} \).

\( V_{\mathrm{A-D}} = V_{\mathrm{A-C}} - 3.53216 \cdot 15 = 89.3424 - 3.53216 \cdot 15 = 89.3424 - 52.9824 = \) 36.3600 V (6 sf) or 36.36 V (4 sf) #.

Looking at node D, the current splits into two branches. The branch current flowing through the 15+10 Ω resistor branch (\( R_{\mathrm{D}} \)) is calculated below.

\( I = \frac{V_{\mathrm{A-D}}}{R_{\mathrm{D}}} = \frac{36.3600}{25} = \) 1.45440 A (6 sf).

The branch current flowing between nodes D and E is calculated using Kirchhoff's Current Law.

\( I_{\mathrm{C-D}} = 3.53216 - 1.45440 = \) 2.07776 A (6 sf).

This current is used to calculate the voltage \( V_{\mathrm{A-E}} \).

\( V_{\mathrm{A-E}} = V_{\mathrm{A-D}} - 2.07776 \cdot 10 = 36.3600 - 2.07776 \cdot 10 = 36.3600 - 20.7776 = \) 15.5824 V (6 sf) or 15.58 V (4 sf) #.