Question

Before leaving the house in the morning, you plop some stew in your slow cooker and turn it on Low. The slow cooker has a 160 Ohm resistor and is plugged into a 120 V outlet. When you come home 8 hours later, how much charge has passed through the slow cooker circuit in that time?

Answer 1

Answer:

Total charge flow through the cooker is 21600 C

Explanation:

As we know that the current flow through the cooker is given by Ohm's law

here it is given as

$V = i R$

$i = \frac{V}{R}$

$i = \frac{120}{160}$

$i = \frac{3}{4} A$

now the charge flow through it is given as

$Q = i t$

total time is t = 8 hours

$Q = \frac{3}{4}(8 \times 60 \times 60)$

$Q = 21600 C$