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11.2Voltage, power and emf
Voltage
The defining equation for resistance, R=VI, can be rearranged in terms of the potential difference V:
V=IR
This says that if there is a current through a conductor that has resistance, i.e. a resistor, then there must be a potential difference across the ends of that resistor. The term voltage is commonly used for potential difference.
Figure 11.11 shows part of a circuit. The current is 5.0 A and the resistance is 15 â¦. The voltage across the resistor is given by V = IR = 5.0 Ã 15 = 75 V. The resistance between B and C is zero, so the voltage across B and C is zero.

Figure 11.11: There is a voltage across points A and B and zero voltage across B and C.

Electric power
We saw that whenever an electric charge q is moved between two points that have a potential difference V, work is done. This work is given by W = qV.
Consider a resistor with a potential difference V across its ends. Since power is the rate of doing work, the power P dissipated in the resistor in moving a charge q across it in time t is:
P=Â work doneÂ Â time takenÂ 
P=qVt
But qt is the current I in the resistor, so the electric power is given by:
P=IV
This power manifests itself in thermal energy and/or work performed by an electrical device.
We can use R=VI to rewrite the formula for power in equivalent ways:
P=IV=RI2=V2R
In a filament lamp electrical energy is transferred to light and thermal energy (Figure 11.12).

Figure 11.12: The metal filament in a light bulb glows as the current passes through it. It is also very hot. This shows that electrical energy is transferred to both thermal energy and light energy.

CHECK YOURSELF 3
The power dissipated in a resistor of constant resistance is 32 W when a potential difference of 12 V is established at its ends.
What is the dissipated power when the potential difference is increased to 15 V?
WORKED EXAMPLE 11.6
A resistor of resistance 12 â¦ has a current of 2.0 A flowing through it. How much energy is generated in the resistor in one minute?
Answer
The power generated in the resistor is:
P=RI2
P=12Ã4=48Â W
Thus, in one minute (60 s) the energy E generated is:
E=48Ã60Â J=2.9Ã103Â J
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Electrical devices are usually rated according to the power they use. A light bulb rated as 60 W at 220 V means that it will dissipate 60 W when a potential difference of 220 V is applied across its ends. If the potential difference across its ends is anything other than 220 V, the power dissipated will be different from 60 W.
WORKED EXAMPLE 11.7
A light bulb rated as 60 W at 220 V has a potential difference of 110 V across its ends. Find the power dissipated in this light bulb.
Answer
Let R be the resistance of the light bulb and P the power we want to find. Assuming R stays constant (so that it is the same when 220 V or 110 V are applied to its ends), we have:
P=1102R and 60=2202R
Dividing the first equation by the second, we find:
P60=11022202
This gives P = 15 W.
EXAM TIP
The power of the light bulb is 60 W only when the voltage across it is 220 V.
If we change the voltage, we will change the power.
Electromotive force (emf)
In an electric circuit, the simplest source of potential difference is a cell or a battery (a battery is a collection of cells). The potential difference is created by chemical reactions inside the cell, Figure 11.13a. The chemicals also create a resistance inside the cell, an internal resistance. In such a cell the source of energy is chemical. In solar cells (photovoltaic cells) the source energy is the solar energy radiated by the sun.
When the cell is connected to an external circuit, Figure 11.13b, electrons flow into the cell from the positive terminal and enter the circuit from the negative terminal. By convention the electric current is taken to be in a direction opposite to the electron flow.

Figure 11.13: a Chemical reactions maintain a potential difference across the terminal of the cell. b When the cell is connected to an external circuit, electrons flow into the battery from the positive terminal and exit into the circuit from the negative terminal. By convention, the current is said to flow in the opposite direction.

A simple mechanical model of a cell involves the following idea. Suppose that in Figure 11.13a you had a âconveyor beltâ that moved electrons from the right terminal of the cell to the left. This would make the right terminal positively charged and the left terminal negatively charged. This would continue until there is enough negative charge on the negative plate to repel any new electrons from arriving there. There is now a potential difference across the terminals. In a real cell, chemical reactions play the role of this imaginary âconveyor beltâ.
KEY POINT
The electromotive force (emf) Îµ (of a cell or battery) is the work done per unit charge in moving charge from one terminal of the battery to the other. The unit of emf is the volt, V. Îµ=emf=Wq
(In batteries, the work done is chemical work. In general, in defining emf, the work done is always nonelectrical.)
By conservation of energy, this work is also equal to the work done W in moving charge q around the circuit. If we divide both numerator and denominator by time, we may also obtain the very convenient fact:
Îµ=emf=PI
So, to find the total power dissipated in a circuit all we have to do is multiply the emf by the current leaving the cell or battery.
This formula is very useful when discussing circuits, as you will see later.
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TEST YOUR UNDERSTANDING
20aWhat is the resistance per unit length of copper wire (resistivity = 1.7 Ã 10â8 Î© m) of radius 4.0 mm?
bA current of 7.5 A is sent down a copper wire of radius 4.0 mm. How much power is dissipated in a length of 150 km of the wire (Figure 11.14)?

Figure 11.14: For question 20.

cWhat is the potential difference at the ends of the 150 km long wire?
21The heating element of an electric kettle dissipates 1.1 kW of power when connected to a source of potential difference 230 V. Calculate the resistance of the heating element.
22The filament of a lamp rated as 60 W at 120 V has resistivity 3.6 Ã 10â7 â¦ m.
aCalculate the resistance of the lamp when it is connected to a source of 120 V.
bThe radius of the filament is 0.042 mm. Determine its length.
cReferring to your answer to part b, suggest how the filament can fit within the light bulb.
23The graph (Figure 11.15) shows how the resistance R of a device varies with applied voltage V for a range of voltages.

Figure 11.15: For question 23.

Sketch a graph to show how the current varies with applied voltage.
24The graph (Figure 11.16) shows how the resistance R of a device varies with applied voltage V.

Figure 11.16: For question 24.

Sketch a graph to show how the current varies with applied voltage.
25Figure 11.17 shows two resistors with a current of 2.0 A in the wire.

Figure 11.17: For question 25.

aCalculate the potential difference across each resistor.
bState the potential difference between points B and C and points A and D.
26A device is rated 60 W at 120 V. What is the current in the device?
27The power dissipated in a resistor is 12 W when the potential difference across it is 6.0 V. What power would be dissipated if the potential difference becomes 9.0 V?
28The power dissipated in a resistor is 4.0 W. What would the dissipated power be if the resistance of the resistor is halved and the potential difference across it stays the same?
29The power dissipated in a resistor is 32 W. What would the dissipated power be if the resistance of the resistor and the potential difference across it are both doubled?
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30In a circuit the emf of the cell is 6.0 V and the current leaving the cell is 2.5 A. What is the total power dissipated in the circuit?
31A charge of 12 C is moved around a circuit. The work done is 48 J. What is the emf of the source?
