Question 1

The objective of this question is to work out the unknown resistance \( R \).

See also: Worksheet 1A, Question 1.

We have the following information:

  • the current flow through the circuit (20 A); and
  • the driving source voltage of the circuit (200 V).


The driving source is depicted as a current source, as the current flow in the circuit is fixed. The current source is drawn as shown so that it produces a positive current in the same direction as the current shown. This circuit may be solved using Ohm's Law:

\( R = \frac{V_{\mathrm{s}}}{I} = \frac{200}{10} = 20 \) i.e. \( R \) = 10 Ω #.


Question 2


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

We have the following information:

  • the resistance of the circuit (45 Ω); and
  • the voltage applied to the circuit (60 V).


This circuit may be solved using Ohm's Law:

\( I = \frac{V}{R} = \frac{60}{45} \) i.e. \( I \) = 1.33 A (to 3 sf) #.


Question 3


Question 3 is similar to Question 2.

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

We have the following information:

  • the resistance of the circuit (95 Ω); and
  • the voltage applied to the circuit (45 V).


This circuit may be solved using Ohm's Law:

\( I = \frac{V}{R} = \frac{45}{95} \) i.e. \( I \) = 0.474 A (to 3 sf) #.


Question 4


The objective of this question is to work out the unknown voltage \( V \).

We have the following information:

  • the resistance of the circuit (35 Ω); and
  • the current flow in the (1.26 A).


This circuit may be solved using Ohm's Law:

\( V = IR = 1.26 \cdot 35 \) i.e. \( V \) = 44.1 V #.


Question 5


The objective of this question is to work out the unknown supply voltage \( V \), similarly to Question 1.

We have the following information:

  • the current flow through the circuit (25 A); and
  • the total resistance of the circuit (9 Ω).


This circuit may be solved using Ohm's Law:

\( V_{\mathrm{s}} = IR = 25 \cdot 9 = 225 \) i.e. \( V_{\mathrm{s}}\) = 225 V #.


Question 6


Series circuits have the following characteristics:

  • The current flow through a series circuit is the same through all the components; and
  • The total voltage drop across a series circuit is the sum of the voltage drops across each individual component; and
  • The total resistance of a set of series resistors is the sum of each individual resistance.


Question 7


The objective of this question is to work out the unknown current \( I \), given the circuit as shown.

We have the following information:

  • the circuit has two 29 Ω resistors in series; and
  • the voltage applied to the circuit is 230 V.


This circuit may be solved using two rules:

  • Series resistance; and
  • Ohm's Law.

The series resistance formula gives us a total resistance (\( R_{\mathrm{TOT}} \)) of \( 29 + 29 \) = 58 Ω.

\( I = \frac{V}{ R_{\mathrm{TOT}}} = \frac{230}{58} \) i.e. \( I \) = 3.97 A (to 3 sf) #.


Question 8


The objective of this question is to work out the unknown current \( I \), given the circuit as shown.

We have the following information:

  • the circuit has a 12 Ω resistor and a 57 Ω resistor in series; and
  • the voltage applied to the circuit is 230 V.


This circuit may be solved using two rules:

  • Series resistance; and
  • Ohm's Law.

The series resistance formula gives us a total resistance (\( R_{\mathrm{TOT}} \)) of \( 12 + 57 \) = 69 Ω.

\( I = \frac{V}{ R_{\mathrm{TOT}}} = \frac{9}{69} \) i.e. \( I \) = 0.130 A (to 3 sf) #.


Question 9


The objective of this question is to work out the unknown current \( I \), given the circuit as shown.

We have the following information:

  • the circuit has the following series resistors: 12 Ω, 27 Ω, 25 Ω, 9 Ω, and 12 Ω; and
  • the voltage applied to the circuit is 150 V.


This circuit may be solved using two rules:

  • Series resistance; and
  • Ohm's Law.

The series resistance formula gives us a total resistance (\( R_{\mathrm{TOT}} \)) of \( 12 + 27 + 25 + 9 + 12 \) = 85 Ω.

The addition method shown above is called binary tree addition. I have broken down the addition into pairs of numbers (to simplify the additions), added them together, then added those numbers pair-wise until I am left with the final sum. This is a particular technique of adding numbers by hand, but any suitable technique will do.

\( I = \frac{V}{ R_{\mathrm{TOT}}} = \frac{150}{85} \) i.e. \( I \) = 1.77 A (to 3 sf) #.


Question 10


State Ohms Law in words:

This question has several possible interpretations, but the answers should cover at least one of these concepts:

  • \( V = I R \) The voltage drop is equal to the product of the current and resistance;
  • \( I = \frac{V}{R} \) The current is the ratio of the voltage drop and the resistance;
  • \( R = \frac{V}{I} \) The resistance is the ratio of the voltage drop and the current.

There are other interpretations of Ohm's Law, but they should cover at least the basic concepts above.

See also:

Last modified: Thursday, 7 May 2020, 10:42 AM