Suppose you have the following circuit – it is similar to a well known circuit called a Wheatstone bridge. Resistors are labeled 1 through 4 for convenience of reference, and the fifth element is a light bulb, which also has some resistance. If any current at all flows through the light bulb, it will glow. You know $R_1 = 150 \Omega$, $R_2=60 \Omega$, and $R_3$ is variable, meaning you can choose its resistance. Part A: You find that when $R_3=250 \Omega$, the light bulb is off. What is $R_4$? Part B: Suppose we set $R_3=500 \Omega$, so that the light bulb glows. We also have the new information that $\Delta V_{\text{bat}} = 20 \text{ V}$. Which direction does conventional current flow through the light bulb? What is the voltage across the light bulb?

• $R_1=150 \Omega$
• $R_2=60 \Omega$
• $R_3=250 \Omega$ for Part A, $R_3=500 \Omega$ for Part B.
• $I_{\text{light}}=0$ for Part A.
• $\Delta V_{\text{bat}} = 20\text{ V}$ for Part B.

• $R_4$
• Direction of conventional current in light bulb for Part B.
• Potential difference across light bulb for Part B.

• The wire has very very small resistance when compared to the other resistors in the circuit.
• The circuit is in a steady state.
• Approximating the battery as a mechanical battery.
• The resistors (including the light bulb) in the circuit are made of Ohmic materials.

• We represent Ohm's Law as

$$\Delta V = IR$$

• We represent the Loop Rule (for potential difference within a closed loop) as

$$\Delta V_1+\Delta V_2+\Delta V_3+\ldots$$

• We represent the Node Rule (for current through a point in the circuit) as

$$I_{\text{in}} = I_{\text{out}}$$

• We represent the situation with diagram given above.

Since there is no current flowing through the light bulb, we also know there is no voltage across it. We can then apply the Loop Rule to the loops highlighted below to find some equivalent voltages: \begin{align*} \Delta V_1 = \Delta V_2, &&&&& \Delta V_3 = \Delta V_4 \end{align*}

A simple application of Ohm's Law changes these equations into \begin{align*} I_1 R_1 = I_2 R_2 &&&&& (1) \\ I_3 R_3 = I_4 R_4 &&&&& (2) \end{align*} We can refer again to there being no current in the light bulb to say more about the current in the rest of the circuit. Since there is no current in that segment, we can use the Node Rule on Nodes A and B to say $I_1=I_3$ and $I_2=I_4$, respectively. When we plug this into the equation (2), we can find that $$R_4 = \frac{I_1}{I_2}R_3$$ It remains to determine the ratio between the two currents. To do this we simply rearrange equation (1) to express the ratio as being between resistors rather than currents. This gives us a final expression for $R_4$: $$R_4 = \frac{R_2}{R_1}R_3 = 100 \Omega$$

To reiterate the new information, we just found that $R_4=100\Omega$. Now, we change Resistor 3 to $R_3=500\Omega$, and the light bulb glows, so we know current is running through it. We also know $\Delta V_{\text{bat}} = 20 \text{ V}$. We want to find which direction conventional current is directed through the light bulb, and the voltage across the light bulb.

The direction of current shouldn't be terribly hard to figure out. If current is directed to the left, it should then add via the Node Rule to the current directed down through Resistor 3. If current is directed to the right, it should then add via the Node Rule to the current directed down through Resistor 4. Before we changed $R_3$, current was zero. Now, we have increased $R_3$. We can imagine that with the increased resistance, current in Resistor 3 should decrease, so it makes sense for current to directed to the right. One could think of this result both as less current in Resistor 3, or more current in Resistor 4. There are also a lot of other arguments that would reach this same conclusion. For clearness, we have drawn the direction of conventional current in each segment below.

To find $\Delta V_{\text{light}}$, we will need to set up some equations using the Loop Rule and Node Rule. We will focus on Nodes C and D, and the Loops highlighted below.

Applying the Node Rule and the Loop Rule, we obtain the following equations: \begin{align*} I &= I_1 + I_2 &(\text{Node C}) \\ I &= I_3 + I_4 &(\text{Node D}) \\ \Delta V_{\text{bat}} &= \Delta V_1 + \Delta V_3 &(\text{Loop 1}) \\ \Delta V_{\text{bat}} &= \Delta V_2 - \Delta V_{\text{light}} + \Delta V_3 &(\text{Loop 2}) \\ \Delta V_1 + \Delta V_{\text{light}} &= \Delta V_2 &(\text{Loop 3}) \\ \Delta V_{\text{light}} + \Delta V_4 &= \Delta V_3 &(\text{Loop 4}) \end{align*} If we combine the equations from Nodes C and D, and apply Ohm's Law, our resistances come into play, which we know. After applying Ohm's Law, what we have is: $$\frac{\Delta V_1}{R_1} + \frac{\Delta V_2}{R_2} = \frac{\Delta V_3}{R_3} + \frac{\Delta V_4}{R_4}$$ Ultimately, we wish to express $\Delta V_{\text{light}}$ in terms of our known resistances, and $\Delta V_{\text{bat}}$. We need to figure out a way to sub in for the unknowns $\Delta V_1$, $\Delta V_2$, $\Delta V_3$, and $\Delta V_4$ in terms of our known constants. Our approach, which is not the only way to solve this problem, is to express some of our unknowns in terms of $\Delta V_3$, and then try to express $\Delta V_3$ in terms of the the potential differences across the battery and light bulb.

First, we use the equations from Loops 1 and 4 to rewrite our most recent result: $$\frac{\Delta V_{\text{light}} - \Delta V_3}{R_1} + \frac{\Delta V_2}{R_2} = \frac{\Delta V_3}{R_3} + \frac{\Delta V_3 - \Delta V_{\text{light}}}{R_4}$$ —

Let's start with resistance. The equivalent resistance for the entire circuit can be found with Ohm's Law – equation (1): $$R_{\text{equiv, circuit}}=\frac{\Delta V_{\text{battery}}}{I_1} = 240\Omega$$

We use $I_1$ since this is the current in the wire connected directly to the battery. We can break this down further to find the equivalent resistance in the chunk of the circuit containing Resistors 2, 3, and 4. This chunk and Resistor 1 are connected in series to form the resistance of the whole circuit, so we can use equation (2) to write: $$R_{\text{equiv, circuit}} = R_1 + R_{\text{equiv, chunk with 2,3,4}}$$ This yields $R_{\text{equiv, chunk with 2,3,4}}=160 \Omega$.

Notice that this “chunk” is actually two parallel pieces of the circuit, starting at Node A and ending at Node B. The two parallel parts are Resistor 2, and then Resistors 3 and 4 together. We can use equation (3) to break down the chunk into these two pieces (we combine Resistors 3 and 4 below using equation (2)): $$\frac{1}{R_{\text{equiv, chunk with 2,3,4}}}= \frac{1}{R_2}+\frac{1}{R_3+R_4}$$ We can plug in what we know and solve for the resistance of Resistor 2: $$R_2=200\Omega$$

Okay, now for the potential differences. It will be useful in the approach we choose to know the current through Resistor 4, which is found from Ohm's Law: $$I_4=\frac{\Delta V_4}{R_4}=10 \text{ mA}$$ A simple application of Node Rule – equation (5) – at Node C should tell us that $I_3 = I_4$. Now, we can reapply Ohm's Law to find the potential difference across Resistor 3: $$\Delta V_3 = I_3R_3 = I_4R_3 = 3 \text{ V}$$

A couple applications of the Loop Rule should help us find the rest of the unknowns. Consider the loop highlighted in the circuit above. The Loop Rule – equation (4) – tells us that if we travel completely around the loop, we should encounter a total potential difference of 0. If we travel along the direction of conventional current (clockwise in our representation), voltage decreases, so $\Delta V_1$, $\Delta V_3$, $\Delta V_4<0$, whereas we have $\Delta V_{\text{bat}}>0$. These four potential differences form a loop, so they should add to 0: $$\Delta V_{\text{bat}}-\Delta V_1 - \Delta V_3 - \Delta V_4 = 0$$ We know enough potential differences to find the voltage across Resistor 1: $$\Delta V_1 = \Delta V_{\text{bat}}-\Delta V_3 - \Delta V_4 =4 \text{ V}$$

Now, consider the (different!) loop highlighted in the circuit above. The Loop Rule tells us that if we travel completely around the loop, we should encounter a total potential difference of 0. We see that current in Resistor 2 runs opposite to the current in the other resistors if we follow the loop in one direction. We have to choose a direction for the application of the Loop Rule. If we go clockwise, voltage increases across Resistor 2, but drops across Resistors 3 and 4. So we write: $$\Delta V_2-\Delta V_3 - \Delta V_4 = 0$$ We know enough potential differences to find the voltage across Resistor 2: $$\Delta V_2 = \Delta V_3+\Delta V_4 = 8 \text{ V}$$ That's all! Note that there are a lot of ways to do this problem, but we chose an approach that showcases the power of knowing equivalent resistance for resistors in parallel, and the power of the Loop Rule. See if you can create a different method for finding the unknowns.