) Draw the graph between voltage dropped across the conductor and current passed through it.
b) Which quantity is obtained from the slope of this graph?
Answers
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According to Ohm's law, the potential difference across a conductor is directly proportional to the current through it.
V=IR
Hence the curve is a straight line passing through origin.
So none of the options is correct.
Electrical power is the rate at which electrical energy is converted in an electric circuit. It calculated as:
P=I⋅V
Power (P) is exactly equal to current (I) multiplied by voltage (V), there is no extra constant of proportionality. The unit of measurement for power is the watt (abbreviated W).
It was James Prescott Joule, not Georg Simon Ohm, who first discovered the mathematical relationship between power dissipation and current through a resistance. This discovery, published in 1841, followed the form of the equation: P=I2R and is properly known as Joule's Law. However, these power equations are so commonly associated with the Ohm's Law equations relating voltage, current, and resistance that they are frequently credited to Ohm.
Equivalent forms
We can use Ohm's Law to show that P=VI is equivalent to P=I2R and P=V2R.
Using V=I⋅R allows us to show:
P=V⋅I=(I⋅R)⋅I Ohm's Law=I2R
Using I=VR allows us to show:
P=V⋅I=V⋅VR Ohm's Law=V2R
WORKED EXAMPLE 8: ELECTRICAL POWER
Given a circuit component that has a voltage of 5 V and a resistance of 2 Ω what is the power dissipated?
Write down what you are given and what you need to find
VRP=5 V=2 Ω=?
Write down an equation for power
The equation for power is:
P=V2R
Solve the problem
P=V2R=(5)2(2)=12,5 W
The power is 12,5 W.
WORKED EXAMPLE 9: ELECTRICAL POWER
Study the circuit diagram below:
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The resistance of the resistor is 15 Ω and the current going through the resistor is 4 A. What is the power for the resistor?
Determine how to approach the problem
We are given the resistance of the resistor and the current passing through it and are asked to calculate the power. We can have verified that:
P=I2R
Solve the problem
We can simply substitute the known values for R and I to solve for P.