Question 1

A marked-up version of the circuit is shown below.


The objective is to find the following quantities:

  • The supply current \( I_{\mathrm{s}} \).
  • The supply voltage \( V_{\mathrm{s}} \).

The following quantities are known:

  • The current through the 100 Ω resistor is 3 A i.e. \( I_{100} = 3 A \).


Solution

The 100 Ω resistor has 3 A flowing through it. This means that the voltage drop across both the 100 Ω resistor and the 25 Ω resistors is 300 V.

The supply current can be calculated, because it is the sum of the 3 A through the 100 Ω resistor, and whatever current flows through the 25 Ω resistor at 300 V drop.

\( I_{\mathrm{s}} = 3 + \frac{300}{25} = 3 + 12 = \) 15 A #.

The supply voltage can now be calculated, using the voltage drop across the 25 Ω and 100 Ω resistors, and the voltage drop across the 27 Ω resistor.

\( V_{\mathrm{s}} = 300 + 15 \cdot 27 = 300 + 405 = \) 705 V #.


Question 2

A marked-up version of the circuit is shown below.


The objective is to find the following quantities:

  • The supply voltage \( V_{\mathrm{s}} \).
  • The supply current \( I_{\mathrm{s}} \).
  • The current through the 15 Ω resistor \( I_{15} \).
  • The unknown resistance \( R_{A} \).

The following quantities are known:

  • The current through the 5 Ω resistor is 2 A i.e. \( I_{5} = 3 A \).
  • The voltage drop across resistor \( R_{A} \) is 70 V.


Solution

The supply current is computable from \( I_{5} \).

The voltage drop across the 5 Ω resistor is 10 V. This is used to compute the supply current, because the parallel 5 Ω, 10 Ω and 8 Ω resistors carry the entire supply current between them. This voltage is 50 V.

\( I_{\mathrm{s}} = 2 + \frac{10}{10} + \frac{10}{8} = 2 + 1 + 1.25 = \) 4.25 A #.

The value of \( R_{A} \) can now be computed, because the voltage and current is known.

\( R_A = \frac{V_A}{I_{\mathrm{s}}} = \frac{70}{4.25} = \) 16.4706 Ω (6 sf) or 16.47 Ω (4 sf) #.

The current through the 15 Ω resistor (\( I_{15} \)) is computed using current division.

\( I_{15} = I_{\mathrm{s}} \cdot \frac{10 \parallel 15}{15} = 4.25 \cdot \frac{6}{15} = \frac{25.5}{15} = \) 1.7 A #.

The supply voltage can now be calculated. Note that the voltage drop across the 15 Ω resistor is 25.5 V.

\( V_{\mathrm{s}} = 10 + 70 + 25.5 = \) 105.5 V #.


Question 3

A marked-up version of the circuit is shown below.


The objective is to find the following quantities:

  • The resistance of resistor \( R_2 \).
  • The resistance of resistor \( R_3 \).
  • The resistance of resistor \( R_4 \).

The following quantities are known:

  • The current through \( R_4 \) is 4 A.
  • The voltage drop across resistor \( R_3 \) is 20 V.
  • The supply current \( I_{\mathrm{s}} \) is 9 A.
  • The supply voltage \( V_{\mathrm{s}} \) is 350 V.


Solution

\( R_4 \) and \( R_5 \) carry the entire supply current between them. This means that the current through \( R_5 \) must be 5 A.

The voltage drop across \( R_4 \) and \( R_5 \) can now be computed using Ohm's Law.

\( V_{R4} = 5 \cdot 36 = \) 180 V.

The voltage drop is used to compute \( R_4 \).

\( R_4 = \frac{180}{4} = \) 45 Ω #.

The value of \( R_3 \) is computed using Ohm's Law, given at 9 A it has a 20 V drop.

\( R_3 = \frac{20}{9} = \) 2.22222 Ω (6 sf) or 2.222 Ω (4 sf) #.

\( R_2 \) can be computed from the voltage drops.

The "remaining" voltage drop across \( R_2 \) is computed using Kirchhoff's Voltage Law.

\( V_{R2} = 350 - 15 \cdot 9 - 20 - 180 = 350 - 135 - 200 = 350 - 335 = \) 15 V.

Ohm's Law is then used to compute \( R_2 \).

\( R_2 = \frac{15}{9} = \) 1.667 Ω (4 sf) #.


Question 4

A marked-up version of the circuit is shown below.


The objective is to find the following quantities:

  • The resistance of resistor \( R_1 \).
  • The resistance of resistor \( R_4 \).
  • The resistance of resistor \( R_5 \).
  • The current through the resistor \( R_4 \): \( I_{R4} \).

The following quantities are known:

  • The supply voltage \( V_{\mathrm{s}} \) is 265 V.
  • The supply current \( I_{\mathrm{s}} \) is 10 A.
  • The voltage drop across parallel resistors \( R_2 \) through \( R_4 \) is 210 V.
  • The voltage drop across resistor \( R_5 \) is 50 V.


Solution

The current through \( R_4 \) is calculated Kirchhoff's Current Law.

\( I_{R4} = 10 - \frac{210}{78} - \frac{210}{91} = \) 5 A.

\( R_4 \) is now calculated using Ohm's Law.

\( R_4 = \frac{210}{5} = \) 42 Ω #.

\( R_5 \) can be computed using Ohm's Law.

\( R_5 = \frac{50}{10} = \) 5 Ω #.

In order to compute \( R_1 \), the voltage drop across it must be known. The voltage drop is calcualted using Kirchhoff's Voltage Law.

\( V_{R4} = 265 - 210 - 50 = \) 5 V #.

The value of \( R_4 \) is computed using Ohm's Law, given at 10 A it has a 5 V drop.

\( R_4 = \frac{5}{10} = \) 0.5 Ω #.


Question 5

A marked-up version of the circuit is shown below.


The objective is to find the following quantities:

  • The resistance of resistor \( R_A \).
  • The resistance of resistor \( R_B \).

The following quantities are known:

  • The supply voltage \( V_{\mathrm{s}} \) is 1000 V.
  • The supply current \( I_{\mathrm{s}} \) is 8 A.
  • The voltage drop across series-parallel resistors 12 Ω, 35 Ω and \( R_B \) is 176 V.

NOTE: The angled wires cross over but don't connect.


Solution

The value of \( R_A \) is calculated using Kirchhoff's Voltage Law. Note that if the voltage drops are known, the exact connection of the resistor network that has that voltage drop is not important.

\( V_{RA} = 1000 - 8 \cdot (17 + 15) - 176 = 1000 - 256 - 176 = 1000 - 432 = \) 568 V.

\( R_A \) is now calculated using Ohm's Law.

\( R_A = \frac{568}{8} = \) 72 Ω #.

The series-parallel network has 176 V drop in total across it.

The voltage drop across the 12 Ω resistor is 96 V. This leaves 80 V across the 35 Ω resistor and \( R_B \).

At 8 A current flow, the parallel combination of these resistors is 10 Ω.

The value of \( R_B \) can be calculated using the parallel resistance formula.

\( \frac{1}{10} = \frac{1}{35} + \frac{1}{R_B} \rightarrow R_B = \frac{1}{\frac{1}{10} - \frac{1}{35}} = \frac{1}{\left( \frac{1}{14} \right)} \) = 14 Ω #.

The current through \( R_B \) is computed using Ohm's Law.

\( I_{RB} = \frac{80}{14} = \) 5.714 A (4 sf) #.


Last modified: Wednesday, 27 May 2020, 3:08 PM