When the lights of an automobile are switched on, an ammeter in series with them

Question

When the lights of an automobile are switched on, an ammeter in series with them reads 9.00 A and a voltmeter connected across them reads 11.60 V. When the electric starting motor is also turned on, the ammeter reading drops to 7.40 A and the lights dim somewhat. If the internal resistance of the battery is 52 m? and that of the ammeter is negligible, what is the emf of the battery

Picture http://spock.physast.uga.edu/res/uga/cwiegert/phys1112/21_circuits/hrw_27p52_mod.problem?symb=uga/song/phys1112/Homework/HW_06.sequence___7___uga/cwiegert/phys1112/21_circuits/hrw_27p52_mod.problem

Explanation / Answer

emf describes the work done per unit charge, and its unit is the volt. If we consider a perfect battery with no internal resistance, then its emf would be equal to the potential difference across the terminals. Real batteries have internal resistance, the terminal voltage is not equal to the emf for a battery. Consider a real battery like a perfect battery with a resistor in series (this resistor is the internal resistance). As we pass from the negative terminal to the positive, the potential increases by an amount E (the potential supplied by the perfect cell) and decreases by Ir (the voltage drop across the internal resistance. So the potential difference across the real cell would be given by: V = E - Ir. Note that E is equal to the open circuit voltage (I = 0). Now lets say there is a load resistor R hooked up to the circuit. The load resistor might be a simple resistive circuit element, or it could be the resistance of an electrical device connected to the battery. The potential difference across the load resistor is V = IR. Using V = E - Ir we get: IR = E - Ir and solving for emf we get: E = IR + Ir E = I(R + r) The internal resistance will need to be given in the question, or enough info must be given to calculate it (such as emf, current and load resistor). An interesting discussion involves power. Solving for current we get: I = E/(R + r) power is given by P = (I^2)(R) = (E^2)(R)/(r+R)^2 Analyzing this equation, we find that the power is at its maximum when R = r.

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