Question

Unit lestCurrent X is 25A and runs for 39 seconds. Current Y is 3.8A and runs for 24 seconds.Which current delivered more charge, and how much more charge did it deliver?

Answer 1

current X delivers more charge,it delivers 0.48 C more of charge

Step-by-step explanation

Current can be calculated using the formula:

`\begin{gathered} I=Qt \\ where\text{ I represents the current} \\ Q\text{ represents charge} \\ t\text{ represents time} \end{gathered}`

so

Step 1

find the charge on situation (current X)

a)let

`\begin{gathered} I\text{ }=25\text{ A} \\ time=39\text{ s} \end{gathered}`

b) now, replace in the formula and solve for Q

`\begin{gathered} I\text{ =Qt} \\ replace \\ 25\text{ A=Q*39 seg} \\ divide\text{ both sides by 39 s} \\ (25A)/(39s)\text{=}(Q*39seg)/(39s) \\ 0.64\text{ C}=Q\text{ } \\ Q_1=0.64C \end{gathered}`

Step 2

find the charge in situation(current Y)

a) let

`\begin{gathered} I=3.8A \\ time=24\text{ s} \end{gathered}`

b) now , replace in the formula and solve for Q

`\begin{gathered} I\text{ =Qt} \\ replace \\ 3.8\text{A=Q*24 seg} \\ divide\text{ both sides by 39 s} \\ (3.8A)/(24s)\text{=}(Q*24seg)/(24s) \\ 0.15\text{ C}=Q\text{ } \\ Q_2=0.1583\text{ }C \end{gathered}`

Step 3

now,

a) compare

`\begin{gathered} Q_1=0.64C \\ Q_2=0.1583\text{C} \\ 0.64>0.1583 \\ therefore \\ Q_1>Q_2 \\ so,\text{ current X deliversmore charge} \end{gathered}`

b)finally, find the difference in the charges

`\begin{gathered} \Delta Q=Q_1-Q_2 \\ \Delta Q=0.64-0.1583 \\ \Delta Q=0.48C \end{gathered}`

therefore, the answer is

current X delivers more charge,it delivers 0.48 C more of charge

I hope this helps you