Video Transcript
The circuit shown contains both
series and parallel combinations of resistors. What is the total current in the
circuit shown? Give your answer to one decimal
place.
Considering our circuit, we see
that it involves four resistors and one cell right here. The positive terminal of that cell
points to the right. So, conventionally, charge will
flow in a clockwise direction around the circuit. After that charge passes through
the first resistor, it will get to this branch point right here. The current will divide up based on
the relative resistances of these two parallel branches, travel through the
branches, and then rejoin. After this, the total or combined
current in the circuit will then pass through this 1.6-ohm resistor and on to the
negative terminal of our cell.
Let’s call the total current in our
circuit 𝐼 sub t. It’s that value that we want to
solve for. We can begin doing this by
recalling Ohm’s law, which tells us that the potential difference across a circuit
equals the current in that circuit multiplied by the total circuit resistance. Dividing both sides of Ohm’s law by
the resistance 𝑅, we find that 𝐼 is equal to 𝑉 divided by 𝑅. So then, if we know the total
potential difference across our circuit 𝑉 and the total resistance of the circuit
𝑅, we can use these values to solve for the total circuit current 𝐼. In our case, the total circuit
current 𝐼 sub t is equal to 5.5 volts, the potential difference supplied by our
cell, divided by the total resistance in the circuit that we’ll call 𝑅 sub t.
To solve for 𝑅 sub t, we’ll need
to combine the resistances of all four of our resistors into one effective
resistance. This process will take a few steps
because as our problem statement notes, here we have resistors connected both in
series and in parallel. Let’s begin by considering the
resistors that are arranged in parallel. In general, if we have exactly two
resistors, 𝑅 one and 𝑅 two, in parallel with one another, then the effective
resistance of these resistors, we’ll call it 𝑅 sub p, is 𝑅 one times 𝑅 two
divided by 𝑅 one plus 𝑅 two.
As we combine resistances in our
circuit to arrive at one total resistance for the circuit, we can say that
effectively this 2.5-ohm resistor and the 3.2-ohm resistor in parallel is the same
as having one single resistor with a resistance value of 2.5 ohms times 3.2 ohms
divided by 2.5 ohms plus 3.2 ohms. Here, we’ve applied our rule for
the equivalent resistance of two resistors in parallel.
Note that we now effectively have
three resistors in our circuit. And these three resistors are all
in series. The equivalent resistance of three
resistors arranged in series with one another is equal to the sum of the individual
resistance values. All this means we now have enough
information to write an expression for the total resistance in our circuit 𝑅 sub
t. By our series resistor addition
rule, we know that 𝑅 sub t equals the resistance of our first resistor, 1.5 ohms,
plus the equivalent resistance of our second resistor plus that of our third
resistor, 1.6 ohms. If we enter this expression for 𝑅
sub t onto our calculator, we get a result of about 4.503 ohms.
We now have a value we can use for
𝑅 sub t in our expression to solve for the total circuit current. 𝐼 sub t equals 5.5 volts divided
by 4.503 and so on ohms. And to one decimal place, this is
equal to 1.2 amperes. To this level of precision, 1.2
amperes is the total current in the circuit.
Let’s now look at part two of our
question.
What is the total power dissipated
by the circuit? Give your answer to one decimal
place.
An electrical circuit dissipates
power by transferring energy away from the circuit over time. For example, power is dissipated
when resistors heat up and transfer heat energy to their surroundings. This part of our question asks us
about the total power dissipated by the circuit. That power is equal in general to
the potential difference across the circuit multiplied by the total circuit
current.
In our case, that potential
difference is 5.5 volts and that current, as we’ve seen, is 5.5 volts divided by the
total circuit resistance. We see that this is equal to the
potential difference provided by the cell squared divided by the total circuit
resistance. Calculating this fraction to one
decimal place, we get a result of 6.7 with units of watts. This is how many joules of energy
are transferred away from our circuit every second. In other words, the power
dissipated by the circuit is 6.7 watts.
