Respuesta :
current X delivers more charge,it delivers 0.48 C more of charge
ExplanationCurrent can be calculated using the formula:
[tex]\begin{gathered} I=Qt \\ where\text{ I represents the current} \\ Q\text{ represents charge} \\ t\text{ represents time} \end{gathered}[/tex]so
Step 1
find the charge on situation (current X)
a)let
[tex]\begin{gathered} I\text{ }=25\text{ A} \\ time=39\text{ s} \end{gathered}[/tex]b) now, replace in the formula and solve for Q
[tex]\begin{gathered} I\text{ =Qt} \\ replace \\ 25\text{ A=Q*39 seg} \\ divide\text{ both sides by 39 s} \\ \frac{25A}{39s}\text{=}\frac{Q\times39seg}{39s} \\ 0.64\text{ C}=Q\text{ } \\ Q_1=0.64C \end{gathered}[/tex]Step 2
find the charge in situation(current Y)
a) let
[tex]\begin{gathered} I=3.8A \\ time=24\text{ s} \end{gathered}[/tex]b) now , replace in the formula and solve for Q
[tex]\begin{gathered} I\text{ =Qt} \\ replace \\ 3.8\text{A=Q*24 seg} \\ divide\text{ both sides by 39 s} \\ \frac{3.8A}{24s}\text{=}\frac{Q\times24seg}{24s} \\ 0.15\text{ C}=Q\text{ } \\ Q_2=0.1583\text{ }C \end{gathered}[/tex]Step 3
now,
a) compare
[tex]\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}[/tex]b)finally, find the difference in the charges
[tex]\begin{gathered} \Delta Q=Q_1-Q_2 \\ \Delta Q=0.64-0.1583 \\ \Delta Q=0.48C \end{gathered}[/tex]therefore, the answer is
current X delivers more charge,it delivers 0.48 C more of charge
I hope this helps you
