Physics · Ch 11 — Electric Current Through Conductors
Electromotive Force (emf)
Electromotive Force (emf)
When charge flows steadily through a conductor, a potential difference must be established -- and continuously MAINTAINED -- across the two ends (terminals) of the conductor, or the flow would simply stop as the charge redistributed itself. A device that maintains this potential difference by doing WORK on the charges passing through it is called an EMF DEVICE, and the work it does per unit charge is called its electromotive force (emf), . Familiar examples of emf devices include dry cells, batteries, solar cells, fuel cells, and generators (Fig. 11.12).
Inside the emf device, positive charge carriers are driven from the region of LOWER potential energy to the region of HIGHER potential energy (i.e. towards the positive terminal, which acts like the 'cathode' inside the device) -- the opposite of what would happen if only electrostatic forces acted. This requires an external, non-electrostatic energy source: chemical energy in an ordinary cell or battery, or the energy of absorbed photons (light) in a solar cell.
If a small charge flows through the circuit's cross-section (including through the emf device itself) in time , the device must do work on this charge to move it from its negative (low-potential) terminal to its positive (high-potential) terminal. This defines the emf:
with SI unit joule/coulomb (J/C), which is the same as the volt.
In an IDEAL emf device, there is no internal resistance opposing the motion of charge carriers inside it, so the device's emf equals the potential difference across its own terminals exactly, whether or not current is flowing. A REAL emf device, however, does have some internal resistance to the motion of charge carriers within it. When a current flows through such a real device (i.e. it is connected in a closed circuit), the terminal voltage falls short of the full emf by the drop across this internal resistance:
(If the device is NOT part of a closed circuit, no current flows, and the terminal voltage simply equals the full emf.) Combining Eq. (11.40) with Ohm's law applied to the external resistor, ,
which rearranges to give the current actually delivered to an external resistance : …
What this figure shows. A closed circuit diagram showing an emf device (drawn with the standard cell/battery symbol of long and short parallel lines) connected to a resistor R by connecting wires. The emf device's positive terminal (+) is marked as being at higher electric potential than its negative terminal (-), with an arrow drawn from the negative to the positive terminal representing the emf itself (distinct from the current-flow arrows in the external wire, which run from the device's positive terminal, through the resis …
Worked out. A resistor network is connected to a 15 V battery with internal resistance 1 Ω: between points A and B there are two 4 Ω resistors in parallel, between B and C a single 1 Ω resistor, and between C and D two 6 Ω resistors in parallel. The worked solution finds the A-B parallel combination (4×4)/(4+4)=2Ω and the C-D parallel combination (6×6)/(6+6)=3Ω, adds these in series with the 1 Ω B-C resistor and the cell's own internal resistance to get the total circuit resistance, applies I = ε/(RT+r) = 15/(6+1) ≈ 2.1 A for the total current, uses symmetry to split this current equally between each pair of parallel resistors (giving 1.05 A through each 4 Ω and each 6 Ω resistor, with the full 2.1 A through the single 1 Ω resistor), and finally computes the voltage drops VBC = I×RBC and VCD = I×RCD, noting that the sum of all voltage drops across the external network is slightly less than the cell's own emf because of the extra drop across the c …