Ohm's
Law:
According
to Ohm’s law, at constant temperature, the voltage difference between two
points on a conductor is proportional to the current passing through it.
Consider a current carrying conductor, having
potential at two points V1 and V2 where V1>V2
According two ohm’s law V1-V2 α
I when temperature is constant.
or V α I
or V = IR (where R is a proportional constant,
which is called resistance of the conductor. The internal property of that
conductor that trends to resist the current flow within it)
Here are two statements that describe the relationship:
1. Given a fixed Resistance, more Voltage causes more Current.
2. Given a fixed Voltage, more Resistance causes less Current.
(Before proceeding I need to remind you that we use Volts for electrical pressure, Amps for electrical current, and Ohms for resistance.)
Ohm's Law is a set of three related equations that encapsulate the two statements above:
V = I * R I = V / R R = V / I
(For historical reasons, we use "V" to represent Voltage, "I" to represent Current, and "R" to represent Resistance.)
1. Given a fixed Resistance, more Voltage causes more Current.
2. Given a fixed Voltage, more Resistance causes less Current.
(Before proceeding I need to remind you that we use Volts for electrical pressure, Amps for electrical current, and Ohms for resistance.)
Ohm's Law is a set of three related equations that encapsulate the two statements above:
V = I * R I = V / R R = V / I
(For historical reasons, we use "V" to represent Voltage, "I" to represent Current, and "R" to represent Resistance.)
Here are some examples using the same flashlight circuit we have
already looked at:
1. Suppose that our bulb has a resistance of 50 ohms and we know that it should be used with a current of 0.03 Amps. What battery voltage is needed?
2. Now suppose that we have a 3 Volt battery and that our bulb has a resistance of 100 ohms. How much current is flowing when the bulb is light? Here are some examples using the same flashlight circuit we have already looked at:
1. Suppose that our bulb has a resistance of 50 ohms and we know that it should be used with a current of 0.03 Amps. What battery voltage is needed?
We know the current and resistance and want to calculate the voltage, so we should choose the first equation ( E = I * R ) then we calculate 50 * 0.03 which equals 1.5 so we need a 1.5V battery.
2. Now suppose that we have a 3 Volt battery and that our bulb has a resistance of 100 ohms. How much current is flowing when the bulb is lit?
1. Suppose that our bulb has a resistance of 50 ohms and we know that it should be used with a current of 0.03 Amps. What battery voltage is needed?
2. Now suppose that we have a 3 Volt battery and that our bulb has a resistance of 100 ohms. How much current is flowing when the bulb is light? Here are some examples using the same flashlight circuit we have already looked at:
1. Suppose that our bulb has a resistance of 50 ohms and we know that it should be used with a current of 0.03 Amps. What battery voltage is needed?
We know the current and resistance and want to calculate the voltage, so we should choose the first equation ( E = I * R ) then we calculate 50 * 0.03 which equals 1.5 so we need a 1.5V battery.
2. Now suppose that we have a 3 Volt battery and that our bulb has a resistance of 100 ohms. How much current is flowing when the bulb is lit?
***Resistance dissipated
Power***
