Question 1

The objective of this question is to work out the supply current \( I_{\mathrm{T}} \).

The circuit is shown below.


Solution - Method 1

Method 1 uses Ohm's Law on the total resistance \( R_{\mathrm{TOT}} \) of the circuit, and from that calculating the current.

\( R_{\mathrm{TOT}} = (21^{-1} + 28^{-1})^{-1} \) = 12 Ω.

The current is computed by using Ohm's Law.

\( I_{\mathrm{s}} = \frac{100}{12} = \frac{25}{3}\) = 8.33 A #.


Solution - Method 2

Method 2 uses Kirchhoff's Current Law to compute the current through the parallel resistors, then sums the currents up to get the total.

  • \( I_{21} = \frac{100}{21} \) A. 
  • \( I_{25} = \frac{100}{28} = \frac{25}{7} \)  A.

The supply current is calculated using \( I_{\mathrm{s}} = I_{21} + I_{28} = \frac{100}{21} + \frac{25}{7} = \frac{25}{3} \) = 8.33 A (3 sf) #.


Question 2

The objective of this question is to work out the following quantities:

  • The current through the 33 Ω resistor \( I_1 \);
  • The current through the 88 Ω resistor \( I_2 \);
  • The current through the 24 Ω resistor \( I_3 \);
  • The total circuit current \( I_{\mathrm{T}} \).

The circuit is shown below.


Solution - Method 1

This circuit can be solved with Ohm's Law and Kirchhoff's Current Law in that order.

First, calculate the resistor currents.

  • \( I_1 = \frac{120}{33} = \frac{40}{11} \) = 3.64 A (3 sf) #.
  • \( I_2 = \frac{120}{88} = \frac{15}{11} \) = 1.36 A (3 sf) #.
  • \( I_3 = \frac{120}{24} \) = 5 A #.

The total current is the sum of these three branch currents.

\( I_{\mathrm{T}} = I_1 + I_2 + I_3 = \frac{40}{11} + \frac{15}{11} + 5 \) = 10 A #.


Solution - Method 2

The resistance as seen by the 120 V voltage source is the total resistance of the three resistors.

\( R_{\mathrm{TOT}} = (33^{-1} + 88^{-1} + 24^{-1})^{-1} \) = 12 Ω.

The supply current is calculated by dividing the supply voltage by \( R_{\mathrm{TOT}} \).

\( I_{\mathrm{T}} = \frac{120}{12} \) = 10 A #.

The resistor currents can also be calculated.

  • \( I_1 = \frac{120}{33} = \frac{40}{11}  \) = 3.64 A (3 sf) #.
  • \( I_2 = \frac{120}{88} = \frac{15}{11}  \) = 1.36 A (3 sf) #.
  • \( I_3 = \frac{120}{24} \) = 5 A #.

Solution - Method 3

The resistance as seen by the 120 V voltage source is the total resistance of the three resistors.

\( R_{\mathrm{TOT}} = (33^{-1} + 88^{-1} + 24^{-1})^{-1} \) = 12 Ω.

The supply current is calculated by dividing the supply voltage by \( R_{\mathrm{TOT}} \).

\( I_{\mathrm{T}} = \frac{120}{12} \) = 10 A #.

The resistor currents can also be calculated using current division \( I = I_{\mathrm{TOT}} \cdot \frac{R_{\mathrm{TOT}}}{R} \).

  • \( I_1 = 10 \cdot \frac{12}{33} = \frac{40}{11}  \) = 3.64 A (3 sf) #.
  • \( I_2 = 10 \cdot \frac{12}{88} = \frac{15}{11}  \) = 1.36 A (3 sf) #.
  • \( I_3 = 10 \cdot \frac{12}{24} \) = 5 A #.


Question 3

The objective of this question is to work out the following quantities:

  • The current through the 91 Ω resistor \( I_1 \);
  • The current through the 78 Ω resistor \( I_2 \);
  • The current through the 21 Ω resistor \( I_3 \);
  • The total circuit current \( I_{\mathrm{T}} \).

The circuit is shown below.


Solution - Method 1

This circuit can be solved with Ohm's Law and Kirchhoff's Current Law in that order.

First, calculate the resistor currents.

  • \( I_1 = \frac{230}{91} \) = 2.53 A (2 dp) #.
  • \( I_2 = \frac{230}{78} = \frac{115}{39} \) = 2.95 A (2 dp) #.
  • \( I_3 = \frac{230}{21} \) = 10.95 A (2 dp) #.

The total current is the sum of these three branch currents.

\( I_{\mathrm{T}} = I_1 + I_2 + I_3 = \frac{230}{91} + \frac{230}{78} + \frac{230}{21}  = \frac{115}{7} \) = 16.43 A (2 dp) #.


Solution - Method 2

The resistance as seen by the 230 V voltage source is the total resistance of the three resistors.

\( R_{\mathrm{TOT}} = (91^{-1} + 78^{-1} + 21^{-1})^{-1} \) = 14 Ω.

The supply current is calculated by dividing the supply voltage by \( R_{\mathrm{TOT}} \).

\( I_{\mathrm{T}} = \frac{230}{14}  = \frac{115}{7} \) = 16.43 A (2dp) #.

The resistor currents can also be calculated.

  • \( I_1 = \frac{230}{91} \) = 2.53 A (2 dp) #.
  • \( I_2 = \frac{230}{78} = \frac{115}{39} \) = 2.95 A (2 dp) #.
  • \( I_3 = \frac{230}{21} \) = 10.95 A (2 dp) #.


Solution - Method 3

The resistance as seen by the 230 V voltage source is the total resistance of the three resistors.

\( R_{\mathrm{TOT}} = (91^{-1} + 78^{-1} + 21^{-1})^{-1} \) = 14 Ω.

The supply current is calculated by dividing the supply voltage by \( R_{\mathrm{TOT}} \).

\( I_{\mathrm{T}} = \frac{230}{14} = \frac{115}{7} \) = 16.43 A (2dp) #.

The resistor currents can also be calculated using current division \( I = I_{\mathrm{TOT}} \cdot \frac{R_{\mathrm{TOT}}}{R} \).

  • \( I_1 = \frac{115}{7} \cdot \frac{14}{91} \) = 2.53 A (2 dp) #.
  • \( I_2 = \frac{115}{7} \cdot \frac{14}{78} \) = 2.95 A (2 dp) #.
  • \( I_3 = \frac{115}{7} \cdot \frac{14}{21} \) = 10.95 A (2 dp) #.

Question 4

The objective of this question is to work out the following quantities:

  • The current through the 5 Ω resistor \( I_1 \);
  • The current through the 12 Ω resistor \( I_2 \);
  • The current through the 20 Ω resistor \( I_3 \);
  • The total circuit current \( I_{\mathrm{T}} \).

The circuit is shown below.


Solution - Method 1

This circuit can be solved with Ohm's Law and Kirchhoff's Current Law in that order.

First, calculate the resistor currents.

  • \( I_1 = \frac{45}{5} \) = 9 A #.
  • \( I_2 = \frac{45}{12} = \frac{15}{12} \) = 3.75 A #.
  • \( I_3 = \frac{45}{20} = \frac{9}{4} \) = 2.25 A #.

The total current is the sum of these three branch currents.

\( I_{\mathrm{T}} = I_1 + I_2 + I_3 = 9 + \frac{15}{12} + \frac{9}{4} \) = 15 A #.


Solution - Method 2

The resistance as seen by the 45 V voltage source is the total resistance of the three resistors.

\( R_{\mathrm{TOT}} = (5^{-1} + 12^{-1} + 20^{-1})^{-1} \) = 3 Ω.

The supply current is calculated by dividing the supply voltage by \( R_{\mathrm{TOT}} \).

\( I_{\mathrm{T}} = \frac{45}{3} \) = 15 A #.

The resistor currents can also be calculated.

  • \( I_1 = \frac{45}{5} \) = 9 A #.
  • \( I_2 = \frac{45}{12} = \frac{15}{4} \) = 3.75 A #.
  • \( I_3 = \frac{45}{20} = \frac{9}{4} \) = 2.25 A #.


Solution - Method 3

The resistance as seen by the 45 V voltage source is the total resistance of the three resistors.

\( R_{\mathrm{TOT}} = (5^{-1} + 12^{-1} + 20^{-1})^{-1} \) = 3 Ω.

The supply current is calculated by dividing the supply voltage by \( R_{\mathrm{TOT}} \).

\( I_{\mathrm{T}} = \frac{45}{3} \) = 15 A #.

The resistor currents can also be calculated using current division \( I = I_{\mathrm{TOT}} \cdot \frac{R_{\mathrm{TOT}}}{R} \).

  • \( I_1 = 15 \cdot \frac{3}{5} \) = 9 A #.
  • \( I_2 = 15 \cdot \frac{3}{12} \) = 3.75 A #.
  • \( I_3 = 15 \cdot \frac{3}{20} \) = 2.25 A #.

Question 5

The objective of this question is to work out the following quantities:

  • The current through the 35 Ω resistor \( I_1 \);
  • The current through the 14 Ω resistor \( I_2 \);
  • The total circuit current \( I_{\mathrm{T}} \).

The circuit is shown below.


NOTE: The order of elements in a parallel circuit does not matter! There are no special steps you need to take to in your calculations just because the voltage source is in the "middle" of the circuit.

Solution - Method 1

This circuit can be solved with Ohm's Law and Kirchhoff's Current Law in that order.

First, calculate the resistor currents.

  • \( I_1 = \frac{25}{35} = \frac{5}{7} \) = 0.71 A (2dp) #.
  • \( I_2 = \frac{25}{14} \) = 1.79 A (2 dp) #.

The total current is the sum of these three branch currents.

\( I_{\mathrm{T}} = I_1 + I_2 = \frac{5}{7} + \frac{25}{14} \) = 2.5 A #.


Solution - Method 2

The resistance as seen by the 25 V voltage source is the total resistance of the two resistors.

\( R_{\mathrm{TOT}} = (35^{-1} + 14^{-1})^{-1} \) = 10 Ω.

The supply current is calculated by dividing the supply voltage by \( R_{\mathrm{TOT}} \).

\( I_{\mathrm{T}} = \frac{25}{10} = \frac{5}{2} \) = 2.5 A #.

The resistor currents can also be calculated.

  • \( I_1 = \frac{25}{35} = \frac{5}{7} \) = 0.71 A (2dp) #.
  • \( I_2 = \frac{25}{14} \) = 1.79 A (2 dp) #.

Solution - Method 3

The resistance as seen by the 25 V voltage source is the total resistance of the two resistors.

\( R_{\mathrm{TOT}} = (35^{-1} + 14^{-1})^{-1} \) = 10 Ω.

The supply current is calculated by dividing the supply voltage by \( R_{\mathrm{TOT}} \).

\( I_{\mathrm{T}} = \frac{25}{10} \) = 2.5 A #.

The resistor currents can also be calculated using current division \( I = I_{\mathrm{TOT}} \cdot \frac{R_{\mathrm{TOT}}}{R} \).

  • \( I_1 = 2.5 \cdot \frac{10}{35} \) = 0.71 A (2 dp) #.
  • \( I_2 = 2.5 \cdot \frac{10}{14} \) = 1.79 A (2 dp) #.

Last modified: Thursday, 7 May 2020, 10:44 AM