Worked examples · Sheet 3

Networks of resistors in series and in parallel

Theory used: Resistors in series Resistors in parallel Voltage divider and current divider Ohm's law

Exercise 1

An ideal source with $E = 12 V$ supplies the resistor $R_{1} = 2.0 Ω$ in series with the parallel combination of the resistors $R_{2} = 6.0 Ω$ and $R_{3} = 3.0 Ω$ . Compute the equivalent resistance, the current delivered by the source and the currents in $R_{2}$ and in $R_{3}$ .

R_{2}

and

R_{3}

are in parallel between the same two nodes; their group is in series with

R_{1}

R_{23} = \frac{R _{2} R _{3}}{R _{2} + R _{3}}, R_{eq} = R_{1} + R_{23}, I = \frac{E}{R _{eq}}, Δ V_{23} = R_{23} I, I_{k} = \frac{Δ V _{23}}{R _{k}}

R_{23} Δ V_{23} = \frac{6.0 \cdot 3.0}{9.0} Ω = 2.0 Ω, R_{eq} = 4.0 Ω, I = \frac{12}{4.0} A = 3.0 A = 2.0 \cdot 3.0 V = 6.0 V, I_{2} = \frac{6.0}{6.0} A = 1.0 A, I_{3} = \frac{6.0}{3.0} A = 2.0 A

The resistors $R_{2}$ and $R_{3}$ are connected between the same two nodes, therefore they are in parallel and are replaced by $R_{23}$ . The resistor $R_{1}$ carries the same current as the group $R_{23}$ , hence it is in series with it, and the equivalent resistance is the sum. Ohm’s law applied to the equivalent resistance gives the current delivered by the source.

Across the parallel combination there is the voltage $R_{23} I = 6.0 V$ , and the rest of the source voltage, $6.0 V$ , drops across $R_{1}$ . Each resistor of the parallel combination has $6.0 V$ across it, and Ohm’s law gives the two currents. As a check, $I_{2} + I_{3} = 3.0 A = I$ , in agreement with conservation of charge at the node; the larger current flows in the resistor with the smaller resistance.

Exercise 2

The parallel combination of the resistors $R_{1} = 4.0 Ω$ and $R_{2} = 12 Ω$ is in series with $R_{3} = 5.0 Ω$ ; the branch thus formed is in parallel with $R_{4} = 8.0 Ω$ , and the whole network is supplied by an ideal source with $E = 24 V$ . Compute the equivalent resistance and the current in each resistor.

The parallel combination of

R_{1}

and

R_{2}

is in series with

R_{3}

, and the branch thus formed is in parallel with

R_{4}

R_{12} = \frac{R _{1} R _{2}}{R _{1} + R _{2}}, R_{123} = R_{12} + R_{3}, R_{eq} = \frac{R _{123} R _{4}}{R _{123} + R _{4}}

R_{12} I Δ V_{12} = \frac{4.0 \cdot 12}{16} Ω = 3.0 Ω, R_{123} = 8.0 Ω, R_{eq} = \frac{8.0 \cdot 8.0}{16} Ω = 4.0 Ω = \frac{24}{4.0} A = 6.0 A, I_{4} = \frac{24}{8.0} A = 3.0 A, I_{3} = \frac{24}{8.0} A = 3.0 A = R_{12} I_{3} = 9.0 V, I_{1} = \frac{9.0}{4.0} A \approx 2.3 A, I_{2} = \frac{9.0}{12} A = 0.75 A

The network is reduced starting from the innermost group. $R_{1}$ and $R_{2}$ are in parallel and are equivalent to $3.0 Ω$ ; this group carries the same current as $R_{3}$ , therefore it is in series with $R_{3}$ , and the branch amounts to $8.0 Ω$ . The branch and $R_{4}$ are connected to the terminals of the source, hence they are in parallel, and two equal resistances in parallel are equivalent to half of one of them, $4.0 Ω$ .

Both branches have the whole voltage $E$ across them and the same resistance, therefore they carry the same current, $3.0 A$ . In the branch, by the voltage divider, the $24 V$ is split into $15 V$ across $R_{3}$ and $9.0 V$ across the parallel combination $R_{12}$ ; Ohm’s law finally gives the currents in $R_{1}$ and $R_{2}$ , which add up to $3.0 A$ . The exact value of $I_{1}$ is $2.25 A$ .

Exercise 3

In a ladder network the terminal $B$ is connected to a common lower line. The resistor $R_{1}$ connects the terminal $A$ to the node $C$ , the resistor $R_{3}$ connects $C$ to the node $D$ ; the resistor $R_{2}$ connects $C$ to the lower line, the resistor $R_{4}$ connects $D$ to the lower line, and the resistors $R_{5}$ and $R_{6}$ , in series, also connect $D$ to the lower line. The voltage $Δ V = 32 V$ is applied between $A$ and $B$ , and the resistors are $R_{1} = 5.0 Ω$ , $R_{2} = 6.0 Ω$ , $R_{3} = 4.0 Ω$ , $R_{4} = 3.0 Ω$ , $R_{5} = 3.0 Ω$ and $R_{6} = 3.0 Ω$ . Compute the equivalent resistance between $A$ and $B$ and the voltage across $R_{6}$ .

Ladder network: it is reduced starting from

R_{5}

and

R_{6}

, the resistors farthest from the terminals

A

and

B

R_{56} = R_{5} + R_{6}, R_{D} = \frac{R _{4} R _{56}}{R _{4} + R _{56}}, R_{C} = \frac{R _{2} ( R _{3} + R _{D} )}{R _{2} + R _{3} + R _{D}}, R_{eq} = R_{1} + R_{C}

R_{56} I Δ V_{D B} Δ V_{6} = 6.0 Ω, R_{D} = \frac{3.0 \cdot 6.0}{9.0} Ω = 2.0 Ω, R_{C} = \frac{6.0 \cdot 6.0}{12} Ω = 3.0 Ω, R_{eq} = 8.0 Ω = \frac{32}{8.0} A = 4.0 A, Δ V_{C B} = R_{C} I = 12 V = \frac{R _{D}}{R _{3} + R _{D}} Δ V_{C B} = \frac{2.0}{6.0} \cdot 12 V = 4.0 V = \frac{R _{6}}{R _{5} + R _{6}} Δ V_{D B} = \frac{3.0}{6.0} \cdot 4.0 V = 2.0 V

The network is reduced starting from the end far from the terminals. $R_{5}$ and $R_{6}$ are in series, because there are no branchings between them; the group $R_{56}$ is connected between the node $D$ and the lower line, like $R_{4}$ , therefore it is in parallel with $R_{4}$ . The equivalent resistor $R_{D}$ is in series with $R_{3}$ , and the group is in parallel with $R_{2}$ ; finally $R_{1}$ is in series with all the rest.

To find the voltage across $R_{6}$ the network is traversed in the reverse direction. The current $I$ enters at $A$ and crosses $R_{1}$ , therefore $Δ V_{C B}$ is the voltage across the equivalent resistance $R_{C}$ . Between $C$ and the lower line, $R_{3}$ and $R_{D}$ are in series, and the voltage divider gives the voltage between $D$ and the lower line; a second divider on $R_{5}$ and $R_{6}$ gives the required voltage. The input voltage is reduced by a factor $16$ , and networks of this type are used precisely as attenuators.