Repeat (Problem)using PSpice.
Reference Problem
A battery charger with a voltage of 12.9 V is used to charge a battery, Figure 4–41(a). The internal resistance of the charger is 0.12 Ω and the voltage of the partially run-down battery is 11.6 V. The equivalent circuit for the charger/battery combination is shown in (b). You reason that, since the two voltages are in opposition, net voltage for the circuit will be 12.9 V - 11.6 V = 1.3 V and, thus, the charging current I will be 1.3 V/0.12 Ω = 10.8 A. Set up the circuit of (b) and use Multisim to verify your conclusion.
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