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Chapter 21: Q14PE (page 775)

Standard automobile batteries have six lead-acid cells in series, creating a total emf of 12.0 V. What is the emf of an individual lead-acid cell?

Short Answer

Expert verified

The emf of an individual lead-acid cell is 2 V.

Step by step solution

01

Principle and formula

Assume emf of individual lead acid cell is . Since, there is series combination of cell which makes battery. Thus, the net emf of a battery is given as,

E = e + e + e + e + e+ e

= 6E

02

Calculation of each cell

When cells are connected in series, the resultant EMF is the algebraic sum of EMF of individual cells. Therefore, here the EMF of individual Lead acid cell is

E=12V6=2V

However actual EMF is 2 V, slightly greater than the value obtained.

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Most popular questions from this chapter

Why can a null measurement be more accurate than one using standard voltmeters and ammeters? What factors limit the accuracy of null measurements?

Two resistors, one having a resistance of900kΩ, are connected in series to produce a total resistance of 0.5MΩ. (a) What is the value of the second resistance? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?

Suppose you are doing a physics lab that asks you to put a resistor into a circuit, but all the resistors supplied have a larger resistance than the requested value. How would you connect the available resistances to attempt to get the smaller value asked for?

(a) What is the resistance of ten275-Ωresistors connected in series? (b) In parallel?

Consider the circuit in Figure21.53, and suppose that the emfs are unknown and the currents are given to beI1=5.00A,I2=3.0A,andI3=-2.00A. (a) Could you find the emfs? (b) What is wrong with the assumptions?
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