Circuit #1!

a) Start from the resistors on the right, and determine whether they are connected in series or parallel.

b) Use Ohm’s Law, and plug in the equivalent resistance.

a) For resistors in parallel, we have:

\begin{equation*}
\frac{1}{R_{eq}’} =\frac{1}{R_2}+\frac{1}{R_3},
\end{equation*}

and since those resistors are connected in parallel, we get:

\begin{equation*}
R_{eq}’ =30 \ \Omega.
\end{equation*}

Then, \(R_1\) and \(R_{eq}’\) are in series. so

\begin{equation*}
R_{eq} = R_{1} + R_{eq}’,
\end{equation*}

or:

\begin{equation*}
R_{eq} = 60 \ \Omega.
\end{equation*}

b) From the equivalent resistance and Ohm’s Law, we get:

\begin{equation*}
I = 2 \ \text{A},
\end{equation*}

which is the same current for \(R_1\). Then, by Ohm’s Law the voltage for \(R_1\) is:

\begin{equation*}
V_1 = 60 \ \text{V}.
\end{equation*}

The voltage for \(R_2\) and \(R_3\) are the same \(V’\). Then \(V’ = V-V_1\):

\begin{equation*}
V’ = 60 \ \text{V}.
\end{equation*}

And the currents are:

\begin{equation*}
I_2 = I_3 = 1 \ \text{A}.
\end{equation*}

For a more detailed explanation of any of these steps, click on “Detailed Solution”.

a) In order to find the equivalent resistance of the circuit, we need to identify groups of resistors that are in series (having the same current), or in parallel (having the same voltage), and apply the equations for the equivalent resistance accordingly. We need to repeat this procedure until we reduce the circuit to a single (equivalent) resistor.

First, notice that \(R_2\) and \(R_3\) share the same wire connections both at the input and the outputs, as illustrated in figure 1.

Figure 1: The resistors \(R_2\) and \(R_3\) share the same connections on both ends.

Because they share the input and output wire connections,  \(R_2\) and \(R_3\)  are subjected to the same voltage and thus are connected in parallel. Hence, the equivalent resistance \(R’_{eq}\) of this pair of resistors is given by

\begin{equation}
\label{Eq1}
\frac{1}{R_{eq}’} =\frac{1}{R_2}+\frac{1}{R_3}.
\end{equation}

We can rewrite this equation as
\begin{equation}
\label{Eq2}
\frac{1}{R_{eq}’} =\frac{\frac{1}{R_2}+\frac{1}{R_3}}{1}.
\end{equation}

If we invert the fractions on both sides, we obtain
\begin{equation}
\label{Eq3}
R_{eq}’ =\frac{1}{\frac{1}{R_2}+\frac{1}{R_3}}
\end{equation}

Now we can insert the numerical values. This gives us

\begin{equation}
\label{val1}
R_{eq}’ =\frac{1}{\frac{1}{60 \ \Omega}+\frac{1}{60 \ \Omega}}=\frac{1}{\frac{2}{60 \ \Omega}}=30 \ \Omega.
\end{equation}

If we replace \(R_2\) and \(R_3\) by their equivalent resistor, we obtain the circuit shown in figure 2.

Figure 2: The equivalent resistor between \(R_2\) and \(R_3\) that were parallel to each other.

Here, both \(R_{1}\) and \(R’_{eq}\) are connected contiguously and the wiring between them does not branch. Hence, \(R_{1}\) and \(R’_{eq}\) have the same current and thus are connected in series. Their equivalent resistance \(R_{eq}\) is then given by the sum of the resistance of both resistors, namely

\begin{equation}
\label{eq3}
R_{eq} = R_{1} + R_{eq}’.
\end{equation}

After inserting the numerical values, we obtain

\begin{equation}
\label{eq4}
R_{eq} = 30 \ \Omega + 30 \ \Omega = 60 \ \Omega,
\end{equation}

and the circuit is reduced to the circuit shown in figure 3.

Figure 3: Final equivalent resistor.

Since we have reduced the circuit to a single resistor and the source, we can say that the equivalent resistance of the circuit is
\begin{equation}
R_{eq} = 60 \ \Omega.
\end{equation}

b) Now that we have that equivalent resistance of the circuit, we can find the total current across the circuit, using Ohm’s law. Next, we will find the voltages and currents of all three resistors. We can do this by analyzing their layout relative to the rest of the circuit, and by applying Ohm’s law sequentially as we expand the circuit back from a single resistor \(R_{eq}\) (as in figure 3) to the initial circuit with all three resistors.

Ohm’s law states that the voltage \(V\) across a resistor equals its resistance \(R\) times the current \(I\) passing through it:

\begin{equation}
\label{eq5}
V = IR.
\end{equation}

In the case of the equivalent resistance (as in figure 3), its voltage is the same as the source voltage since it is connected to it from both ends without any other resistor in-between. Hence, Ohm’s law in this case can be written as

\begin{equation}
\label{eq6}
V = IR_{eq},
\end{equation}

where \(V=120 \) V is the source voltage, \(I\) is the total current going through the circuit, and \(R_{eq}\) is the equivalent resistance that we found in part a). If we divide by \(R_{eq}\) on both sides, we obtain

\begin{equation}
\label{eq7}
I = \frac{V}{R_{eq}}.
\end{equation}

After inserting numerical values, we get
\begin{equation}
\label{eq71}
I = \frac{120 \ \text{V}}{60 \ \Omega} = 2 \frac{\text{V}}{\ \Omega} = 2 \ \text{A}.
\end{equation}

Now, notice in figure 2 that this is the same current that passes through \(R_{1}\) and \(R’_{eq}\). Hence,
\begin{equation}
\label{eq8}
I_1 = I = 2\ \text{A},
\end{equation}

and we can apply Ohm’s law to find the voltage across \(R_{1}\):
\begin{equation}
\label{eq9}
V_1 = I_1R_1.
\end{equation}

After inserting the numerical values, we obtain

\begin{equation}
\label{eq10}
V_1 = 2\ \text{A}\cdot30\ \Omega = 60\ \text{V}.
\end{equation}

Now, the voltage across the entire circuit must equal the voltage of the source. By applying this to the circuit in figure 2, we can say that the voltage \(V_1\) across \(R_{1}\) plus the voltage V’ across \(R’_{eq}\) must equal the voltage of the source. We can write this as:
\begin{equation}
\label{eq11}
V_1 + V’ = V.
\end{equation}

If we subtract \(V_1\) from both sides of this equation, we get
\begin{equation}
\label{eq12}
V’ = V-V_1,
\end{equation}

and after inserting the numerical values we get
\begin{equation}
\label{eq13}
V’ = 120 \ \text{V} – 60 \ \text{V} = 60 \ \text{V}.
\end{equation}

Now, recall from part a) and figure 2 that \(R’_{eq}\) is the equivalent resistance of resistors \(R_2\) and \(R_3\), which have the same voltage (i.e. they are connected in parallel). Hence, V’ must be the same as the voltage across \(R_2\) and \(R_3\), that is
\begin{equation}
\label{eq14}
V’ = V_2 = V_3.
\end{equation}

After inserting the corresponding numbers, we obtain
\begin{equation}
\label{eq15}
V_2 = 60 \ \text{V},
\end{equation}

and
\begin{equation}
\label{eq16}
V_3 = 60 \ \text{V}.
\end{equation}

Finally, we can apply Ohm’s law to find the current through \(R_2\) and \(R_3\). For \(R_2\) we get
\begin{equation}
\label{eq17}
V_2 = I_2 R_2.
\end{equation}

After dividing by \(R_2\), we obtain
\begin{equation}
\label{eq18}
I_2 = \frac{V_2}{R_2}.
\end{equation}

Similarly, for \(R_3\) we get
\begin{equation}
\label{eq19}
I_3 = \frac{V_3}{R_3},
\end{equation}

Insert the numerical values to obtain
\begin{equation}
\label{eq20}
I_2 = \frac{60 \ \text{V}}{60\ \Omega} = 1 \ \text{A},
\end{equation}

and
\begin{equation}
\label{eq21}
I_3 = \frac{60 \ \text{V}}{60\ \Omega} = 1 \ \text{A}.
\end{equation}

As a validation of the results, notice that \(I_2 + I_3 = 2\) A, which is the value of \(I_1\). This is expected since the current must be conserved as it splits from \(R_{1}\) into the branches containing \(R_{2}\) and \(R_{3}\). Furthermore, \(I_2 = I_3\), as expected since the current \(I_1\) must be equally split between both resistors given that they have the same resistance.

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