Question 24.8: A battery that has an emf E1 = 9 V and internal resistance r......
A battery that has an emf \mathcal{E}_{1} = 9 V and internal resistance r_{1} = 0.02 Ω is connected to a second battery of \mathcal{E}_{2} = 12 V and r_{2} = 0.04 Ω, such that their like terminals are connected, see Fig. 24.13. Find the current in the circuit and the terminal voltage across each battery
The two batteries are oppositely directed around the circuit. Since \mathcal{E}_{2} > \mathcal{E}_{1} , then the net emf \mathcal{E}_{net} in this circuit will be in the counterclockwise direction, i.e.:
{\mathcal{E}}_{\mathrm{net}}={\mathcal{E}}_{2}-{\mathcal{E}}_{1}=12-9\,\mathrm{V}=3\,\mathrm{V} (Counterclockwise direction)
Consequently, the current I in this circuit will also be in the counterclockwise direction as indicated in Fig. 24.13. This current is opposite to the discharging current that the \mathcal{E}_{1} = 9 V battery should produce when connected to circuits containing only resistors. Actually, this current will charge the \mathcal{E}_{1} = 9 V battery.
The total resistance of this circuit is only due to the presence of the internal resistances r_{1} and r_{2} of the two batteries. Therefore, Eq. 24.28 gives us the value of the current as follows:
I={\frac{\mathcal{E}}{R+r}} (24.28)
I={\frac{{\mathcal{E}}_{2}-{\mathcal{E}}_{1}}{r_{1}+r_{2}}}={\frac{12\,{\mathrm{V}}-9\,{\mathrm{V}}}{0.02\,{\Omega}+0.04\,{\Omega}}}={\frac{3\,{\mathrm{V}}}{0.06\,{\Omega}}}=50\,{\mathrm{A}}Depending on the direction of the current in each battery, the terminal voltages
across the batteries are:
ΔV = V_{b} − V_{a} = \mathcal{E}_{1} + Ir_{1} = 9 V + (50 A)(0.02 Ω)
= 10 V (Gain from a to b)
ΔV = V_{b′} − V_{a′} = \mathcal{E}_{2} − I_{r2} = 12 V − (50 A)(0.04 Ω)
= 10 V (Drop from a′ to b′)