Download Lesson T5D - Ohm`s Law

OCS
Spring 2016
Radio Electronics
Lesson 4 - Ohm’s Law (T5D)
Page 1
Pastor Shafer
Almost anyone who knows much of anything about electricity can recite Ohm’s law. In January
of 1781, Henry Cavendish used Leyden jars filled with saline solution to measure current by
paying attention to how strong of a shock he felt in his body when he used himself to complete
the circuit (there is a better way to do this!).
German physicist Georg Ohm experimented with “resistance” in the years 1825 and 1826. We
get the name of the standard unit of resistance from his name – the Ohm. The Greek letter
Omega “ Ω ” is often used as the symbol for the ohm. An ohm is the resistance between two
points on a conductor when one volt across those two points produces a current flow of one
ampere. Energy is dissipated by a resistor in the form of heat.
Ohm’s Law
To remember Ohm’s law, remember the acronym “EAR” (E = Amps (A) times R (resistance).
To convert the equation to other forms, use simple algebra (as seen above).
T5D01 The formula is used to calculate current in a circuit is, Current (I) equals voltage
(E) divided by resistance (R).
T5D02 The formula is used to calculate voltage in a circuit is Voltage (E) equals current
(I) multipliedby resistance (R).
T5D03 The formula is used to calculate resistance in a circuit is Resistance (R) equals
voltage (E)divided by current (I).
Consider a circuit with a battery and a resistor. The current
flowing through the circuit can be calculated by taking the
voltage of the battery and dividing by the resistance rating of the
resistor. For example, if the battery is a 90 volt battery, and the
resistance is 30 ohms, the current is 90/30 or 3 amperes. If you
know any two of the three variables (current, voltage, and
resistance), you can find the other one. Since power (measured
in Watts) is also based on voltage and current (P = E x I) you can calculate the power being
dissipated by the resistor. In this example, since E=90 and I=3, power = 270 Watts.
The equations for power (P), in terms of voltage (E) or current (I), and resistance (R), are:
P = E 2/R
and
P=I2R
(since P = EI) and E = IR and I = E/R
OCS
Spring 2016
Page 2
Pastor Shafer
Radio Electronics
Lesson 4 - Ohm’s Law (T5D)
Here are the problems you will probably see on the test.
T5D04 The resistance of a circuit in which a current of 3 amperes flows through a resistor
connected to 90 volts is 30 ohms.
R = E / I or R = 90 / 3 or R = 30 ohms
T5D05 The resistance in a circuit for which the applied voltage is 12 volts and the current
flow is 1.5 amperes is 8 ohms.
R = E / I or R = 12 / 1.5 or R = 8 ohms
T5D06 The resistance of a circuit that draws 4 amperes from a 12-volt source is 3 ohms.
R = E / I or R = 12 / 4 or R = 3 ohms
I
(current in Amperes)
=E
(voltage in Volts) Multiplied by
R
(resistance in Ohms)
T5D07 What is the current flow in a circuit with an applied voltage of 1 20 volts and a
resistance of80 ohms is 1.5 amperes.
I = E / R or I = 120 / 80 or I = 1.5 Amperes
T5D08 The current flowing through a 100-ohm resistor connected across 200 volts is 2
amperes.
I = E / R or I = 200 / 100 or I = 2 Amperes
T5D09 The current flowing through a 24-ohm resistor connected across 240 volts is 10.
Amperes.
I = E / R or I = 240 / 24 or I = 10 Amperes
T5D10 The voltage across a 2-ohm resistor if a current of 0.5 amperes flows through it is
1 volt.
E = I x R or I = .5 x 2 or E = 1 Volts
T5D11 The voltage across a 10-ohm resistor if a current of 1 ampere flows through it is 10
volts.
E = I x R or I = 1 x 10 or E = 10 Volts
T5D12 The voltage across a 10-ohm resistor if a current of 2 amperes flows through it is
20 volts.
E = I x R or I = 2 x 10 or E = 20 Volts
All of this is nothing more than simple algebra, if you know the equation. You are solving one
equation with one unknown (in a single loop circuit).