Worksheet 15A - Basic Series-Parallel Calculations - Solutions and Commentary

Introduction

These circuits are worked out using series-parallel circuit calculations. The maximum simplification is that there are no nested parallel circuits at each step.

In other words:

• Calculations like \( R_1 + (R_2 + R_3) \parallel (R_4 + R_5) \parallel (R_6 + R_7) \) are done in a single step.
• Calculations like \( R_1 + (R_2 + R_3) \parallel (R_4 + R_5 \parallel (R_6 + R_7)) \) are broken down into multiple steps, due to the nested parallel calculation.

These rules are followed throughout these solutions.

Question 1

The marked up schematic is shown below.

The objective is to calculate the load current \( I_{\mathrm{L}} \).

Solution

The load current is given by Ohm's Law

\( I_{\mathrm{L}} = \frac{V_{\mathrm{s}}}{R_{\mathrm{L}}} \)

where \( I_{\mathrm{L}} \) is the load current, \( V_{\mathrm{s}} \) is the source/supply voltage, and \( R_{\mathrm{L}} \) is the total load resistance.

The load resistance is calculated using standard series-parallel workings.

\( R_{\mathrm{L}} = 21 \parallel 28 + 28 = 12 + 28 = \) 40 Ω.

The load current is now calculated by Ohm's Law.

\( I_{\mathrm{L}} = \frac{V_{\mathrm{s}}}{R_{\mathrm{L}}} = \frac{150}{40} = \) 3.75 A #.

Question 2

The marked up schematic is shown below.

The objective is to calculate the load current \( I_{\mathrm{L}} \).

Solution

The load current is given by Ohm's Law.

\( I_{\mathrm{L}} = \frac{V_{\mathrm{s}}}{R_{\mathrm{L}}} \)

where \( I_{\mathrm{L}} \) is the load current, \( V_{\mathrm{s}} \) is the source/supply voltage, and \( R_{\mathrm{L}} \) is the total load resistance.

The load resistance is calculated using standard series-parallel workings.

\( R_{\mathrm{L}} = 7 \parallel 42 + 10 + 6 \parallel 6 \parallel 6 = 6 + 10 + 2 = \) 18 Ω.

The load current is now calculated by Ohm's Law.

\( I_{\mathrm{L}} = \frac{V_{\mathrm{s}}}{R_{\mathrm{L}}} = \frac{36}{18} = \) 2 A #.

Question 3

The marked up schematic is shown below.

The objective is to calculate the supply voltage \( V_{\mathrm{s}} \).

Solution

The supply voltage is given by Ohm's Law.

\( V_{\mathrm{s}} = I_{\mathrm{L}} R_{\mathrm{L}} \)

where \( V_{\mathrm{s}} \) is the source/supply voltage, \( I_{\mathrm{L}} \) is the load current, and \( R_{\mathrm{L}} \) is the total load resistance.

The load resistance is calculated using standard series-parallel workings.

\( R_{\mathrm{L}} = 5 + 8 + 4 + 21 \parallel 28 = 17 + 12 = \) 29 Ω.

The supply voltage is now calculated by Ohm's Law.

\( V_{\mathrm{s}} = I_{\mathrm{L}} R_{\mathrm{L}} = 10 \cdot 29 = \) 290 V #.

Question 4

The marked up schematic is shown below.

The objective is to calculate:

• the current through \( R_2 \): \( I_{R2} \);
• the current through \( R_3 \): \( I_{R3} \);
• the current through \( R_4 \): \( I_{R4} \).

NOTE: This circuit as shown is inconsistent. It is not possible to have the combination of supply voltage, load current and resistances shown. This is because \( R_5 \) should be \( \frac{10}{3} \) Ω, not 3.33 Ω. The effects of this will be discussed in the solutions.

Solution - Current Division

The unknowns may be computed by means of current division.

The formula for current division is

\( I_{Rn} = I_{\mathrm{T}} \cdot \frac{R_{\mathrm{T}}}{R_n} \)

where \( I_{Rn} \) is the current through resistor \( R_n \), \( I_{\mathrm{T}} \) is the total current through all the parallel resistors, and \( R_{\mathrm{T}} \) is the total parallel resistance of all the parallel resistors.

The total resistance is calculated using the standard parallel resistance workings.

\( R_{\mathrm{T}} = 5 \parallel 20 \parallel 12 = \) 3 Ω.

The currents through the individual resistances can now be calculated using the current division formula. Note that the total current is 12 A.

\( I_{R2} = 12 \cdot \frac{3}{5} = \) 7.2 A #.

\( I_{R3} = 12 \cdot \frac{3}{20} = \) 1.8 A #.

\( I_{R4} = 12 \cdot \frac{3}{12} = \) 3 A #.

The total current is 12 A, which matches the total given.

Solution - Ohm's Law

The unknowns may be computed by means of Ohm's Law.

The currents through \( R_2 \) through \( R_4 \) may be found using the voltage drop across these resistors. This voltage drop will be called \( V_{R2} \).

\( V_{R2} = V_{\mathrm{s}} - V_{R1} - V_{R5} = 100 - 12 \cdot (2 + 3.33) = 100 - 63.96 = \) 36.04 V (4 sf).

The currents through the individual resistances can now be calculated using Ohm's Law.

\( I_{R2} = \frac{36.04}{5} = \) 7.208 A # (4 sf).

\( I_{R3} = \frac{36.04}{20} = \) 1.802 A # (4 sf).

\( I_{R4} = \frac{36.04}{12} = \) 3.003 A # (4 sf).

NOTE: The currents do not quite "balance". This is because the 3.33 Ω resistor is not specified precisely enough to produce 12 A load current to 4 sf. In this case, the method using current division is more precise, as it uses the 12 A load current directly. It does not use the imprecise 3.33 Ω resistor.

Question 5

The marked up schematic is shown below.

The objective is to calculate the unknown resistance \( R_6 \).

NOTE: The wires that cross at an angle are not connected.

Solution

While \( R_6 \) is not known, the total resistance \( R_{\mathrm{T}} \) can be calculated from the supply voltage and current.

\( R_{\mathrm{T}} = \frac{V_{\mathrm{s}}}{R_{\mathrm{L}}} = \frac{75}{2} = \) 37.5 Ω.

The total resistance is 37.5 Ω. The unknown resistance \( R_6 \) is calculated by subtracting the known resistances from the total.

\( R_6 = R_{\mathrm{T}} - 8 - 6 - 10 \parallel 90 - 7 = 37.5 - 21 - 9 = \) 7.5 Ω #.