Worksheet 8A - Series and Parallel problems - Words - Solutions and Commentary

Question 1

Three resistors are connected in parallel and then this combination is connected in series with a 10 Ohm resistor.

When 500 Volts is applied to the circuit, 25 Amps flows in the 10 Ohm series resistor.

If 2 of the parallel resistors are 28 Ohms each then what is the value of the unknown resistor, and what is the current through it and the voltage drop across it?

Solution

The things we need to know to compute the unknown resistor (I will call it \( R_5 \)) are:

• the unknown resistor is in parallel with two 28 Ω resistors;
• the total current flow through all three parallel resistors is 25 A;
• the 25 A flows from a 500 V source and a 10 Ω series resistor.

The two 28 Ω resistors will be treated as one 14 Ω resistance to ease the rest of the calculations.

The equation for the current through \( R_5 \) is as follows.

\( I_{R5} = 10 \cdot \frac{14 \parallel R_5}{R_5} \)

The problem is that we have two unknowns, but one equation.

A second equation is obtained from the source voltage, and the voltage drop across the 10 Ω series resistor.

\( V_{R5} = 500 - 25 \cdot 10 = \) 250 V #.

So, R_5 has 250 V across it.

The current through the unknown resistance is as follows.

\( I_{R5} = 25 - \frac{250}{14} = \frac{50}{7} = \) 7.14286 A ( 6 sf) or 7.143 A (4 sf) #.

The resistance of \( R_5 \) is calculated using Ohm's Law.

\( R_5 = \frac{V_{R5}}{I_{R5}} = \frac{250}{7.14286} = \) 35 Ω (6 sf) #.

Question 2

A heating element is in two sections, each of 45 Ohms.

Find the current drawn from the supply ( 230V ) when the sections are connected in:

1. series; and
2. parallel.

Solution

a) Series Resistances

When the resistors are connected in series, the total load resistance is 90 Ω.

The load current is given by Ohm's Law.

\( I_{\mathrm{L}} = \frac{V_{\mathrm{s}}}{R_{\mathrm{L}}} = \frac{230}{90} = \) 2.556 A (4 sf) #.

b) Parallel Resistances

When the resistors are connected in parallel, the total load resistance is 22.5 Ω.

The load current is given by Ohm's Law.

\( I_{\mathrm{L}} = \frac{V_{\mathrm{s}}}{R_{\mathrm{L}}} = \frac{230}{22.5} = \) 10.22 A (4 sf) #.

Question 3

Two cables having resistances of 0.5 Ohms and 0.8 Ohms carry between them a current of 30 Amps.

What is the current in each cable and the Voltage Drop that occurs when the current is flowing?

Solution

The current through each cable is evaluated by current division.

The parallel resistance of the cables is \( 0.5 \parallel 0.8 = \) 0.307692 Ω (6 sf).

The voltage drop can be found from the parallel resistance and the total current.

\( V_{\mathrm{D}} = I_{\mathrm{L}} R_{\mathrm{C}} = 30 \cdot 0.307692 = \) 9.231 V (4 sf) #.

The current through the 0.5 Ω resistance cable is evaluated by current division.

\( I_{0.5} = I_{\mathrm{L}} \cdot \frac{0.307692}{0.5} = 30 \cdot 0.615384 = \) 18.46 A (4 sf) #.

The current through the 0.8 Ω resistance cable is evaluated by current division.

\( I_{0.8} = I_{\mathrm{L}} \cdot \frac{0.307692}{0.8} = 30 \cdot 0.384615 = \) 11.54 A (4 sf) #.

The current through the "other" cable (in this case the 0.8 Ω resistance cable) could also be found via Kirchhoff's Current Law.

\( I_{0.8} = I_{\mathrm{L}} - I_{0.5} = 30 - 18.46 = \) 11.54 A (4 sf) #.

Question 4

A cable carries a current of 45 Amps and when that happens a voltage drop of 21 Volts occurs.

What would be the resistance value of a cable connected in parallel with the first to reduce the voltage drop to 5% of 230 Volts.

Solution

The current current cable resistance is calculated below.

\( R_1 = \frac{21}{45} = \frac{7}{15} = \) 0.466667 Ω (6 sf).

The target voltage drop \( V_{\mathrm{D}} \) is calculated below.

\( V_{\mathrm{D}} = 230 \cdot 0.05 = \) 11.5 V.

Assuming the load current stays the same, the target cable resistance is calculated.

\( R_{\mathrm{C}} = \frac{11.5}{45} = \frac{23}{90} = \) 0.255556 Ω (6 sf).

A parallel cable must be found with a resistance that reduces the total cable resistance from 0.466667 Ω to 0.255556 Ω.

This can be solved for using the parallel resistance formula with fractions.

\( \frac{1}{R_{\mathrm{C}}} = \frac{1}{R_1} + \frac{1}{R_2} \rightarrow \frac{1}{R_2} = \frac{1}{R_1} - \frac{1}{R_{\mathrm{C}}} \rightarrow R_2 = \frac{1}{\frac{1}{R_1} - \frac{1}{R_{\mathrm{C}}}} \)

\( R_2 = \frac{1}{\frac{1}{R_1} - \frac{1}{R_{\mathrm{C}}}} = \frac{1}{\frac{90}{23} - \frac{15}{7}} = \frac{1}{\left( \frac{285}{161} \right)} = \frac{161}{285} = \) 0.5649 Ω (4 sf) #.

Alternatively, it can be solved numerically.

\( R_2 = \frac{1}{\frac{1}{R_1} - \frac{1}{R_{\mathrm{C}}}} = \frac{1}{3.91304 - 2.14286} = \frac{1}{1.77018} = \) 0.5649 Ω (4 sf) #.

Question 5

The specification for a wire to be used in a wire wound resistor says that its resistance is 0.6 Ohms per meter. How many meters would be required to give:

1. a resistance of 3.7 Ohms, and;
2. 0.2 Ohms, given that the wire must be at least a meter long to reach from one end of the resistor to the other?

Solution

The wire has a value \( r \) of 0.6 Ω.m^-1.

The formula for resistance of \( n \) equal-length strands in parallel is:

\( R = \frac{r L}{n} \)

where \( R \) is the resistance, \( L \) is the length of each strand of resistance wire, and \( n \) is the number of strands.

In all cases, the number of strands is minimised.

a) The 3.7 Ω Resistor

No minimum length is specified. The number if strands is assumed to be 1 (i.e. \( n = 1 \)).

\( R = \frac{r L}{n} \rightarrow L = \frac{n R}{r} \)

Plugging in the known values...

\( L = \frac{1 \cdot 3.7}{0.8} = \) 6.167 m (4 sf) #.

b) The 0.2 Ω Resistor on the 1 m Long Former

The length formula in terms of the required resistance and \( r \) value is given below.

\( L = \frac{n \cdot 0.2}{0.6} = 0.333333 n \).

The length must be at least 1 m.

It appears this is achieved for any \( n \geq 3 \).

In other words, 3 1 m lengths in parallel will achieve a resistance of 0.2 Ω #.

Question 6

A circuit consists of three resistors initially connected in series and then in parallel.

What in general terms occurs to the following, given the changes listed.

Solution

A table of answers is given below.