Question

A 12 V car battery can send a total charge of 84 A.h (ampere-hours) through a circuit, from one terminal to the other.
(a) How many coulombs of charge does this represent?
(b) How much work will be done by the battery to move this entire charge across the circuit?

• A. $3.0\times {10}^5$C, $3.6\times {10}^6$J
• B. $1.5\times {10}^5$C, $1.8\times {10}^6$J
• C. $0.75\times {10}^5$C, $0.9\times {10}^6$J
• D. $1.0\times {10}^5$C, $1.0\times {10}^6$J

Answer 1

The correct option is A $3.0\times {10}^5$C, $3.6\times {10}^6$J

(a) An ampere is as defined a coulomb per second, so

$84A.h=(84\frac{C}{s}.h)$

1 hour = 3600 s, so

$\Rightarrow 84A.h=\left(84\frac{C}{s}.3600s\right)$
$\Rightarrow 84A.h=(84\times 3600)$ C
$\Rightarrow 84A.h=3.0\times {10}^5$ C.

(b) The potential differnce between two points is defined as the the work done to move a unit positive charge across those two points.

$\Delta V=\frac{W}{Q}$

Given Potential difference across the circuit = $12V$

Total charge to be moved across the circuit = $3.0\times {10}^{5}C$

Total work done by the battery to move the chrage across the circuit is
$W=q\Delta V$
$=(3.0\times {10}^5)\left(12V\right)$
$=3.6\times {10}^{6}J.$