Question 24.7: A device is connected to a battery that has an emf ε = 9 V a......
A device is connected to a battery that has an emf \mathcal{E} = 9 V and internal resistance r = 0.02 Ω. Find the current in the circuit and the terminal voltage of the battery when the device is a: (a) light bulb that has a resistance R = 4 Ω, see Fig. 24.12a. (b) conducting wire having zero resistance, i.e. the battery is short circuited by this conductor, see Fig. 24.12b.
(a) Equation 24.28 gives us the value of the current as:
I={\frac{\mathcal{E}}{R+r}} (24.28)
={\frac{9\mathrm{\ V}}{4\ \Omega+0.02\ {\Omega}}}=2.24\,\mathrm{A}From Eq. 24.25, the terminal voltage of the battery will be given by:
ΔV = \mathcal{E} − Ir (ΔV = \mathcal{E} for an open-circuit) (24.25)
ΔV = \mathcal{E} − Ir = 9 V − (2.24 A)(0.02 Ω) = 8.96 V
(b) When we use a conducting wire, it is as if we have a device of R = 0. This results in a current and terminal voltage of the battery as follows:
ΔV = \mathcal{E} − Ir = 9 V − (450 A)(0.02 Ω) = 0
Such large values for the current I would result in a very quick depletion of the battery as all of its stored energy would be quickly transferred to the conducting wire in the form of heat energy. The term “short circuit” is applied to such cases, and they can cause fire or burns.