Name:_____MrsDunn_________
Ohm’s Law Practise Worksheet
Part A
(Reference - http://mrsganske.wikispaces.com/2012-13+Physical+Science+Term+2A )
Show all four steps for each problem
STEP 1 - WRITE EQUATION
STEP 2 – RE-ARRANGE EQUATION IF NEEDED
STEP 3 - SUBSTITUTE IN THE EQUATION
STEP 4 – SOLVE
1.
Find the current through a 12-ohm resistive circuit when 24 volts is applied.
I = V/R 
= 24/12
= 2A
2. Find the resistance of a circuit that draws 0.06 amperes with 12 volts applied.
R=V/I
= 12/0.06
= 200 Ω or 200 Ohm
3. Find the applied voltage of a circuit that draws 0.2 amperes through a 4800-ohm resistance.
V=IR
=0.2*4800
= 960V
4. Find the applied voltage of a telephone circuit that draws 0.017amperes through a resistance of 15,000
ohms.
V = IR
= 0.017x 15000
= 255V
5. A 20-volt relay has a coil resistance of 200 ohms. How much current does it draw?
I = V/R
= 20/200
= 0.1A
6. A series circuit has 1200-ohms of total resistance with 12 V as the power supply. What is the total current
of this circuit?
I = V/R
= 12/1200
= 0.01A
7. A transformer is connected to 120 volts. Find the current if the resistance is 480-ohms?
I = V/R
= 120/480
= 0.25A 
8. A resistive load of 600-ohms is connected to a 24 V power supply. Find the current through the resistor.
I = V/R
= 24/600
= 0.04A
9. A circuit consists of a 12 V battery connected across a single resistor. If the current in the circuit is 3 A,
calculate the size of the resistor.
R = V/I
= 12/3
= 4 Ohm
10. What is the increase of current when 15 V is applied to 10000-ohm rheostat, which is adjusted to 1000ohm value?
I = V/R
I1 = 15/10000
= 0.015A
= 15mA
I2 = 15/1000
= 0.0015A
=1.5mA
Difference or Increase = 15-1.5 = 13.5mA = 0.0135A
Part B
(Reference - http://www.allaboutcircuits.com/worksheets/ohm_law.html)
1. Explain, step by step, how to calculate the amount of current (I) that will go through the resistor
in this circuit: Ohm’s Law
I=V/R
= 12/470
= 0.0255A
= 25.5mA
2. What is the value of this resistor, in ohms (Ω)?
R=V/I
= 12.3/4.556
= 2699.7 Ohm
= 2K7 Ohm
3. A common saying about electricity is that it always takes the path of least resistance." Explain
how this proverb relates to the following circuit, where electric current from the battery
encounters two alternate paths, one being less resistive than the other:
I1 = V/R = 10/250 = 40mA
I2 = V/R = 10/800 = 12.5mA
I1-I2 = 40 - 12.5 = 37.5mA
The current through the 250Ohm resistor is more. 
4. What would happen if a wire having no resistance at all (0 Ω) were connected directly across the
terminals of a 6-volt battery? How much current would result, according to Ohm's Law?
I = V/R
=6/0

Heat up, Short circuit