A 12 V car battery dies not so much because its voltage drops but because chemical reactions increase its internal resistance. A good battery connected with jumper cables can both start the engine and recharge the dead battery. Consider the automotive circuit of FIGURE P28.64. How much current is the dead battery alone able to drive through the starter motor?
What are the battery current Ibat and the potential difference V₁ - V₂ between points 1 and 2 when the switch in FIGURE P28.55 is (a) open and (b) closed?

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Key Concepts
Ohm's Law
Series and Parallel Circuits
Voltage Drop
A lightbulb is in series with a 2.0 Ω resistor. The lightbulb dissipates 10 W when this series circuit is connected to a 9.0 V battery. What is the current through the lightbulb? There are two possible answers; give both of them.
Suppose you have resistors 2.5 Ω, 3.5 Ω, and 4.5 Ω and a 100 V power supply. What is the ratio of the total power delivered to the resistors if they are connected in parallel to the total power delivered if they are connected in series?
What is the current through the 10 Ω resistor in FIGURE P28.61? Is the current from left to right or right to left?
Load resistor R is attached to a battery of emf and internal resistance r. For what value of the resistance R, in terms of ∈ and r, will the power dissipated by the load resistor be a maximum?
A circuit you're building needs an ammeter that goes from 0 mA to a full-scale reading of 50 mA. Unfortunately, the only ammeter in the storeroom goes from 0 μA to a full-scale reading of only 500 μA. Fortunately, you've just finished a physics class, and you realize that you can make this ammeter work by putting a resistor in parallel with it, as shown in FIGURE P28.56. You've measured that the resistance of the ammeter is 50.0 Ω, not the 0 Ω of an ideal ammeter. What is the effective resistance of your ammeter?
