Figure 27-7b is the same circuit except that point b is now directly connected
to ground. Thus, the potential there is defined to be V b ϭ 0. Equation 27-11 now
tells us that the potential at a is V a ϭ Ϫ8.0 V.
Power, Potential, and Emf
When a battery or some other type of emf device does work on the charge carriers to establish a current i, the device transfers energy from its source of energy (such as the chemical source in a battery) to the charge carriers. Because a
real emf device has an internal resistance r, it also transfers energy to internal
thermal energy via resistive dissipation (Module 26-5). Let us relate these
transfers.
The net rate P of energy transfer from the emf device to the charge carriers is
given by Eq. 26-26:
P ϭ iV,
( 2 7 - 1 4 )
where V is the potential across the terminals of the emf device. From Eq. 27-13,
we can substitute V ϭ Ᏹ Ϫ ir into Eq. 27-14 to find
P ϭ i(Ᏹ Ϫ ir) ϭ iᏱ Ϫ i
2
r.
( 2 7 - 1 5 )
From Eq. 26-27, we recognize the term i
2
r in Eq. 27-15 as the rate P r of energy
transfer to thermal energy within the emf device:
P r ϭ i
2
r (internal dissipation rate).
(27-16)
Then the term iᏱ in Eq. 27-15 must be the rate P emf at which the emf device
transfers energy both to the charge carriers and to internal thermal energy.
Thus,
P emf ϭ iᏱ (power of emf device).
(27-17)
If a battery is being recharged, with a “wrong way” current through it, the
energy transfer is then from the charge carriers to the battery — both to the
battery’s chemical energy and to the energy dissipated in the internal resistance r.
The rate of change of the chemical energy is given by Eq. 27-17, the rate of dissipation is given by Eq. 27-16, and the rate at which the carriers supply energy is
given by Eq. 27-14.
779
27-1 SI NG LE-LOOP CI RCU ITS
Figure 27-7 (a) Point a is directly connected to ground. (b) Point b is directly connected to
ground.
R = 4.0 Ω
i
r = 2.0 Ω
= 12 V
i
a
b +
–
(a)
R = 4.0 Ω
i
r = 2.0 Ω
= 12 V
i
a
b +
–
(b)
Ground is taken
to be zero potential.
Checkpoint 3
A battery has an emf of 12 V and an internal resistance of 2 ⍀. Is the terminal-toterminal potential difference greater than, less than, or equal to 12 V if the current in
the battery is (a) from the negative to the positive terminal, (b) from the positive to
the negative terminal, and (c) zero?
to ground. Thus, the potential there is defined to be V b ϭ 0. Equation 27-11 now
tells us that the potential at a is V a ϭ Ϫ8.0 V.
Power, Potential, and Emf
When a battery or some other type of emf device does work on the charge carriers to establish a current i, the device transfers energy from its source of energy (such as the chemical source in a battery) to the charge carriers. Because a
real emf device has an internal resistance r, it also transfers energy to internal
thermal energy via resistive dissipation (Module 26-5). Let us relate these
transfers.
The net rate P of energy transfer from the emf device to the charge carriers is
given by Eq. 26-26:
P ϭ iV,
( 2 7 - 1 4 )
where V is the potential across the terminals of the emf device. From Eq. 27-13,
we can substitute V ϭ Ᏹ Ϫ ir into Eq. 27-14 to find
P ϭ i(Ᏹ Ϫ ir) ϭ iᏱ Ϫ i
2
r.
( 2 7 - 1 5 )
From Eq. 26-27, we recognize the term i
2
r in Eq. 27-15 as the rate P r of energy
transfer to thermal energy within the emf device:
P r ϭ i
2
r (internal dissipation rate).
(27-16)
Then the term iᏱ in Eq. 27-15 must be the rate P emf at which the emf device
transfers energy both to the charge carriers and to internal thermal energy.
Thus,
P emf ϭ iᏱ (power of emf device).
(27-17)
If a battery is being recharged, with a “wrong way” current through it, the
energy transfer is then from the charge carriers to the battery — both to the
battery’s chemical energy and to the energy dissipated in the internal resistance r.
The rate of change of the chemical energy is given by Eq. 27-17, the rate of dissipation is given by Eq. 27-16, and the rate at which the carriers supply energy is
given by Eq. 27-14.
779
27-1 SI NG LE-LOOP CI RCU ITS
Figure 27-7 (a) Point a is directly connected to ground. (b) Point b is directly connected to
ground.
R = 4.0 Ω
i
r = 2.0 Ω
= 12 V
i
a
b +
–
(a)
R = 4.0 Ω
i
r = 2.0 Ω
= 12 V
i
a
b +
–
(b)
Ground is taken
to be zero potential.
Checkpoint 3
A battery has an emf of 12 V and an internal resistance of 2 ⍀. Is the terminal-toterminal potential difference greater than, less than, or equal to 12 V if the current in
the battery is (a) from the negative to the positive terminal, (b) from the positive to
the negative terminal, and (c) zero?