Question 1

The objective of this question is to work out the unknown current \( I \).

We have the following information:

  • the resistances in the circuit; and
  • the driving source voltage of the circuit (500 V).


Before calculating the current, the total resistances must be calculated.

\( R_{\mathrm{TOT}} = \sum R = 15 + 15 + 20 = 50 \), i.e. \( R_{\mathrm{TOT}} \) = 50 Ω. The symbol \( \sum \) (Greek capital letter sigma) means 'sum'.

This circuit may now be solved using Ohm's Law:

\( I = \frac{V_{\mathrm{s}}}{R_{\mathrm{TOT}}} = \frac{500}{50} = 1 \) i.e. \( I \) = 10 A #.


Question 2


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

  • The voltage drop across the 25 Ω resistor \( V_{25} \);
  • The voltage drop across the 2.5 Ω resistor \( V_{2.5} \);
  • The voltage drop across the 10.5 Ω resistor \( V_{10.5} \);
  • The supply voltage \( V_{\mathrm{s}} \).

We have the following information:

  • the driving source current is 5.5 A.


Three methods of solving this circuit are shown below.

Method 1: Kirchhoff's Voltage Law (KVL) Method

The KVL method relies on the supply voltage being the sum of the voltage drops across each resistor i.e. \( V_{\mathrm{s}} = V_{25} + V_{2.5} + V_{10.5} \).

The resistor voltage drops are calculated using \( V = I R \)

  • \( V_{25} = 5.5 \cdot 25 \) = 137.5 V #;
  • \( V_{2.5} = 5.5 \cdot 2.5 \) = 13.75 V #;
  • \( V_{10.5} = 5.5 \cdot 10.5 \) = 57.75 V #;

The supply voltage is calculated by adding up the individual resistor voltage drops.

\( V_{\mathrm{s}} = \sum V = V_{25} + V_{2.5} + V_{10.5} = 137.5 + 13.75 + 57.75 \) = 209 V #.

This method is the most direct, since every result is one of the "final" results. There are no "wasted" calculations.

Method 2: Series Resistance Method

The series resistance method is similar to the KVL method except it uses the total resistance to calculate the supply voltage. The voltages across the individual resistors are calculated the same way:

  • \( V_{25} = 5.5 \cdot 25 \) = 137.5 V #;
  • \( V_{2.5} = 5.5 \cdot 2.5 \) = 13.75 V #;
  • \( V_{10.5} = 5.5 \cdot 10.5 \) = 57.75 V #;

The supply voltage is calculated using the total resistance \( R_{\mathrm{TOT}} \).

\( R_{\mathrm{TOT}} = \sum R = 25 + 2.5 + 10.5 = 38 \), i.e. \( R_{\mathrm{TOT}} \) = 38 Ω.

The supply voltage can then be calculated using Ohm's Law.

\( V_{\mathrm{s}} = I R_{\mathrm{TOT}}  = 5.5 \cdot 38 \) = 209 V #.

This result is the same as with the KVL method, which is good.

The series resistance method is not as direct as the KVL method as it requires the calculation of the total resistance, which is not one of the final answers.

Method 3: Voltage Divider Method

This method uses the voltage divider equation to calculate the resistor voltages.

To perform this method, the total resistance \( R_{\mathrm{TOT}} \) must be determined first.

\( R_{\mathrm{TOT}} = \sum R = 25 + 2.5 + 10.5 = 38 \), i.e. \( R_{\mathrm{TOT}} \) = 38 Ω.

The supply voltage also needs to be calculated, to determine the total voltage across the series resistors.

\( V_{\mathrm{s}} = I R_{\mathrm{TOT}} = 5.5 \cdot 38 \) = 209 V #.

The voltages can then be calculated using the voltage divider equation \( V = V_{\mathrm{s}} = V \cdot \frac{R}{R_{\mathrm{TOT}}} \).

  • \( V_{25} = 209 \cdot \frac{25}{38} \) = 137.5 V #;
  • \( V_{2.5} = 209 \cdot \frac{2.5}{38} \) = 13.75 V #;
  • \( V_{10.5} = 209 \cdot \frac{10.5}{38} \) = 57.75 V #;

This method is the least favoured for this circuit, since the advantage of the voltage divider method is that you don't need the current. However, we know the current already, so the advantage is nullified.


Question 3


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

  • The voltage drop across the 3 Ω resistor \( V_{3} \) (written as \( VD_3 \));
  • The supply current \( I_{\mathrm{s}} \).

We have the following information:

  • the supply voltage is 35 V.


Method 1

Method 1 relies on finding the supply current \( I_{\mathrm{s}} \) first, then using it to calculate the voltage drops.

The total resistance \( R_{\mathrm{TOT}} \) is required to calculate the supply current.

\( R_{\mathrm{TOT}} = \sum R = 3 + 5 + 12 = 20 \), i.e. \( R_{\mathrm{TOT}} \) = 20 Ω.

The supply current \( I_{\mathrm{s}} \) is calculated using Ohm's Law.

\( I_{\mathrm{s}} = \frac{35}{20} \) = 1.75 A #.

The voltage drop across the 3 Ω resistor is calculated using Ohm's Law.

 \( V_{3} = 3 \cdot 1.75 \) = 5.25 V #.

Method 2

This method uses the voltage divider equation to compute \( V_{3} \) first, then uses \( V_{3} \) to compute \( I_{\mathrm{s}} \).

Computing the voltage across the 3 Ω resistor requires calculating \( R_{\mathrm{TOT}} \).

\( R_{\mathrm{TOT}} = \sum R = 3 + 5 + 12 = 20 \), i.e. \( R_{\mathrm{TOT}} \) = 20 Ω.

The voltage across the 3 Ω resistor can then be calculated.

\( V_{3} = 35 \cdot \frac{3}{20} \) = 5.25 V #.

The supply current \( I_{\mathrm{s}} \) can be calculated from Ohm's Law using \( V_{3} \). You could also use the 12 Ω or 5 Ω resistors to calculate \( I_{\mathrm{s}} \), but there is no point doing these calculations because (i) we already have the information we need; and (ii) we are not interested in their voltage drops.

\( I_{\mathrm{s}} = \frac{V_{3}}{3} = \frac{5.25}{3} \) = 1.75 A #.

This method is slightly less direct than method 1, but still gets the correct answers.


Question 4


This question is similar to Question 2.

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

  • The voltage drop across the 37 Ω resistor \( V_{37} \) (written as \( VD_{37} \));
  • The voltage drop across the 25 Ω resistor \( V_{25} \) (written as \( VD_{25} \));
  • The supply current \( I_{\mathrm{s}} \).

We have the following information:

  • the supply voltage is 200 V.


Method 1

Method 1 relies on finding the supply current \( I_{\mathrm{s}} \) first, then using it to calculate the voltage drops.

The total resistance \( R_{\mathrm{TOT}} \) is required to calculate the supply current.

\( R_{\mathrm{TOT}} = \sum R = 37 + 25 + 10.73 = 72.73 \), i.e. \( R_{\mathrm{TOT}} \) = 72.73 Ω.

The supply current \( I_{\mathrm{s}} \) is calculated using Ohm's Law.

\( I_{\mathrm{s}} = \frac{200}{72.73} \) = 2.750 A (4 sf) #.

The voltage drops across the 37 Ω and 25 Ω resistor are calculated using Ohm's Law.

  • \( V_{37} = 37 \cdot 2.75 \) = 101.75 V (2 dp) #.
  • \( V_{25} = 25 \cdot 2.75 \) = 68.75 V (2 dp) #.

Method 2

This method uses the voltage divider equation to compute \( V_{37} \) and \( V_{25} \) first, computes \( I_{\mathrm{s}} \).

Computing the voltage across the resistors requires calculating \( R_{\mathrm{TOT}} \).

\( R_{\mathrm{TOT}} = \sum R = 37 + 25 + 10.73 = 72.73 \), i.e. \( R_{\mathrm{TOT}} \) = 72.73 Ω.

The voltages across the 37 Ω and 25 Ω resistors can then be calculated.

  • \( V_{37} = 200 \cdot \frac{37}{72.73} \) = 101.75 V (2dp) #.
  • \( V_{25} = 200 \cdot \frac{25}{72.73} \) = 68.75 V (2dp) #.

The supply current \( I_{\mathrm{s}} \) can be calculated from Ohm's Law using \( V_{37} \) or \( V_{25} \).

NOTE: You only need to do one of these calculations. You could also do the calculations using the voltage drop across the 10.73 Ω resistor, but there is no point doing this calculation, because (i) we already have the information we need; and (ii) we are not interested in the voltage drop across the 10.73 Ω resistor.

  • \( I_{\mathrm{s}} = \frac{V_{37}}{37} = \frac{101.75}{37} \) = 2.75 A #.
  • \( I_{\mathrm{s}} = \frac{V_{25}}{25} = \frac{68.75}{25} \) = 2.75 A #.

This method is slightly less direct than method 1, but still gets the correct answers.


Question 5


This question is similar to Question 2.

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

  • The voltage drop across the 16 Ω resistor \( V_{16} \) (written as \( VD_{16} \));
  • The voltage drop across the 25 Ω resistor \( V_{25} \) (written as \( VD_{25} \));
  • The voltage drop across the 45 Ω resistor \( V_{45} \) (written as \( VD_{45} \));
  • The supply voltage \( V_{\mathrm{s}} \).

We have the following information:

  • the load current is 2.5 A.


Three methods of solving this circuit are shown below.

Method 1: Kirchhoff's Voltage Law (KVL) Method

The KVL method relies on the supply voltage being the sum of the voltage drops across each resistor i.e. \( V_{\mathrm{s}} = V_{16} + V_{25} + V_{45} \).

The resistor voltage drops are calculated using \( V = I R \)

  • \( V_{16} = 2.5 \cdot 16 \) = 40 V #;
  • \( V_{25} = 2.5 \cdot 25 \) = 62.5 V #;
  • \( V_{45} = 2.5 \cdot 45 \) = 112.5 V #;

The supply voltage is calculated by adding up the individual resistor voltage drops.

\( V_{\mathrm{s}} = \sum V = V_{16} + V_{25} + V_{45} = 40 + 62.5 + 112.5 \) = 215 V #.

This method is the most direct, since every result is one of the "final" results. There are no "wasted" calculations.

Method 2: Series Resistance Method

The series resistance method is similar to the KVL method except it uses the total resistance to calculate the supply voltage. The voltages across the individual resistors are calculated the same way:

  • \( V_{16} = 2.5 \cdot 16 \) = 40 V #;
  • \( V_{25} = 2.5 \cdot 25 \) = 62.5 V #;
  • \( V_{45} = 2.5 \cdot 45 \) = 112.5 V #;

The supply voltage is calculated using the total resistance \( R_{\mathrm{TOT}} \).

\( R_{\mathrm{TOT}} = \sum R = 16 + 25 + 45 = 86 \), i.e. \( R_{\mathrm{TOT}} \) = 86 Ω.

The supply voltage can then be calculated using Ohm's Law.

\( V_{\mathrm{s}} = I R_{\mathrm{TOT}}  = 2.5 \cdot 86 \) = 215 V #.

This result is the same as with the KVL method, which is good.

The series resistance method is not as direct as the KVL method as it requires the calculation of the total resistance, which is not one of the final answers.

Method 3: Voltage Divider Method

This method uses the voltage divider equation to calculate the resistor voltages.

To perform this method, the total resistance \( R_{\mathrm{TOT}} \) must be determined first.

\( R_{\mathrm{TOT}} = \sum R = 16 + 25 + 45 = 86 \), i.e. \( R_{\mathrm{TOT}} \) = 86 Ω.

The supply voltage also needs to be calculated, to determine the total voltage across the series resistors.

\( V_{\mathrm{s}} = I R_{\mathrm{TOT}} = 2.5 \cdot 86 \) = 215 V #.

The voltages can then be calculated using the voltage divider equation \( V = V_{\mathrm{s}} = V \cdot \frac{R}{R_{\mathrm{TOT}}} \).

  • \( V_{25} = 215 \cdot \frac{16}{86} \) = 40 V #;
  • \( V_{2.5} = 215 \cdot \frac{25}{86} \) = 62.5 V #;
  • \( V_{10.5} = 215 \cdot \frac{40}{86} \) = 112.5 V #;

This method is the least favoured for this circuit, since the advantage of the voltage divider method is that you don't need the current. However, we know the current already, so the advantage is nullified.

Last modified: Thursday, 7 May 2020, 10:43 AM