Video Transcript
Which of the following is the
correct formula for the current through a point in a circuit? πΌ represents the current, π
represents the amount of charge, and π‘ represents time. (A) πΌ is equal to π divided by
π‘. (B) πΌ is equal to π multiplied by
π‘. (C) πΌ is equal to π‘ divided by
π. (D) π is equal to πΌ multiplied by
π‘ squared.
In this question, we want to find
the formula for the current through a point in a circuit. We will begin by recalling the
definition for electric current.
Electric current is the rate at
which electric charge passes a point in a circuit. When an amount of charge π passes
a point in the circuit in a time π‘, the current πΌ is given by πΌ is equal to π
divided by π‘. This corresponds to option (A). We can confirm that this equation
is correct by looking at the units for each of these quantities.
Units are an important part of
every physics problem. Not only do they communicate the
exact type of quantity we are working with, but they can also include clues to help
solve the problem. The SI unit for charge is the
coulomb, which is denoted by a capital C. The SI unit for time is the second,
which is denoted by a lowercase s. The SI unit for current is the
ampere, which is denoted by a capital A. One ampere is equal to one coulomb
per second.
Looking at the formula πΌ is equal
to π divided by π‘ in option (A), we see that the left-hand side has units of
amperes and the right-hand side has units of coulombs divided by seconds. We know that the ampere is equal to
one coulomb per second. So we can see that the units match
on both sides of this equation.
Now letβs think about the units of
the other options weβve been presented with. Looking at the formula πΌ is equal
to π multiplied by π‘ in option (B), we see that the left-hand side has units of
amperes and the right-hand side has units of coulombs multiplied by seconds. Since amperes are equal to coulombs
per second, we can substitute this in on the left-hand side. This gives us that coulombs divided
by seconds equals coulombs multiplied by seconds. We can now see that the units on
both sides of this equation do not match. This means that this formula is
incorrect, and so option (B) is incorrect.
Looking at the formula πΌ is equal
to π‘ divided by π in option (C), we see that the left-hand side has units of
amperes and the right-hand side has units of seconds divided by coulombs. Replacing amperes with coulombs per
second, we see that the left-hand side becomes coulombs divided by seconds. Again, we see that the units on
both sides of this equation do not match. This means that this formula is
incorrect, so option (C) is incorrect.
Looking at the formula π is equal
to πΌ multiplied by π‘ squared in option (D), we see that the left-hand side has
units of coulombs and the right-hand side has units of amperes multiplied by seconds
squared. Replacing amperes with coulombs per
second and simplifying, we find that the right-hand side becomes coulombs multiplied
by seconds. Again, we see that the units on
both sides of this equation do not match. This means that this formula is
incorrect, and option (D) is incorrect.
And so we can confirm that option
(A) is indeed the correct answer. πΌ is equal to π divided by π‘ is
the correct formula for the current through a point in a circuit.