A particular 12 V car battery can send a total charge of 110 A·h (ampere-hours) through a circuit, from one terminal to the other. (a) How many coulombs of charge does this represent? (Hint: See ) (b) If this entire charge undergoes a change in electric potential 12 V, how much energy is involved?

(a) 3.96 x 10⁵C

(b) 4.752 x 10⁶ J

(a) The given charge (Q) is 110 A·h (ampere hour)

Converting this to A·s (ampere second) gives the number of coulombs the charge represents. This is done as follows;

=> Q = 110A·h

=> Q = 110 x 1A x 1h          [1 hour = 3600 seconds]

=> Q = 110 x A x 3600s

=> Q = 396000A·s

=> Q = 3.96 x 10⁵A·s = 3.96 x 10⁵C

Therefore, the number of coulombs of charge is 3.96 x 10⁵C

(b) The energy (E) involved in the process is given by;

E = Q x V           (i)

Where;

Q = magnitude of the charge = 3.96 x 10⁵C

V = electric potential = 12V

Substitute these values into equation (i) as follows;

E = 3.96 x 10⁵ x 12

E = 47.52 x 10⁵ J

E = 4.752 x 10⁶ J

Therefore, the amount of energy involved is 4.752 x 10⁶ J

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