Suppose you have the following circuit. Resistors are labeled 1 through 4 for convenience of reference. You know that the circuit contains a 12-Volt battery, and $R_1=80 \Omega$, $R_2=200 \Omega$, $I_1 = 50 \text{ mA}$, and $\Delta V_3=3 \text{ V}$. What is the resistance of and power dissipated through Resistor 4?

• $\Delta V_{\text{bat}} = 12\text{ V}$
• $R_1=80 \Omega$
• $R_2=200 \Omega$
• $I_1 = 50 \text{ mA}$
• $\Delta V_3 = 3\text{ V}$

• $R_4$, $P_4$.

• 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 in the circuit are made of Ohmic materials.

• We represent Ohm's Law as

\begin{align*} \Delta V = IR &&&&&& (1) \end{align*}

• We represent power dissipated across a potential as

\begin{align*} P = I\Delta V &&&&&& (2) \end{align*}

• We represent the equivalent resistance of multiple resistors arranged in series as

\begin{align*} R_{\text{equiv, series}} = R_1+R_2+R_3+\ldots &&&&&& (3) \end{align*}

• We represent the equivalent resistance of multiple resistors arranged in series as

\begin{align*} \frac{1}{R_{\text{equiv, parallel}}}= \frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\ldots &&&&&& (4) \end{align*}

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

\begin{align*} \Delta V_1+\Delta V_2+\Delta V_3+\ldots = 0 &&&&&& (5) \end{align*}

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

\begin{align*} I_{\text{in}} = I_{\text{out}} &&&&&& (6) \end{align*}

• We represent the situation with diagram given above.

If we wish to find information about Resistor 4, we may it useful to find $R_3$, and then use what we know about parallel resistors to find $R_4$. In order to get to this point, we can take advantage of the Loop Rule and Node Rule to find potential differences and currents in the circuit. Let's start with the Loop Rule. Consider the loop in the circuit highlighted below. The Loop Rule – equation (5) – 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 $\Delta V_{battery}>0$. These four potential differences form a loop, so they should add to 0: $$\Delta V_{bat}-\Delta V_1 - \Delta V_3 - \Delta V_4 = 0$$ We can use Ohm's Law to find $\Delta V_1=I_1R_1=4 \text{ V}$, and now we have enough potential differences to find the voltage across Resistor 4: $$\Delta V_4 = \Delta V_{bat}-\Delta V_1 - \Delta V_3 =5 \text{ V}$$

Consider the (different!) loop in the circuit highlighted below. 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 just look at the current in the highlighted loop. Notice that we can't just follow the direction of conventional current in our loop, so 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 can solve this using known potential differences for $\Delta V_2=8 \text{ V}$. Now, we know the potential difference across every resistor. Next, we will try to find the resistances themselves. We know $R_1$ and $R_2$. We also know from Ohm's Law – equation (1) –that the equivalent resistance for the entire circuit should be $$R_{\text{equiv, circuit}}=\frac{\Delta V_{\text{battery}}}{I_1} = 240\Omega$$ 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 (3) 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 (4) to break down the chunk into these two pieces:

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We can use the Loop Rule – equation (4) – to find the potential difference across these two resistors. The potential difference across the battery has opposite sign as the differences across the resistors, if we consider the circuit as a loop of individual differences. We write: $$\Delta V_{bat} = \Delta V_1 + \Delta V_2 + \Delta V_3$$ Since we know $\Delta V_{bat}$ and $\Delta V_3$, we can plug in and solve: $\Delta V_1 + \Delta V_2 = 6 \text{ V}$.

We can find current through the circuit using equations (1) and (2). We can write the power dissipated through Resistor 1 as $$P_1={I_1}^2R_1$$ Since we know $P_1$ and $R_1$, we can plug in and solve for $I_1=\sqrt{P_1/R_1}=0.1 \text{ A}$. Recall that the Node Rule tells us that the current is the same everywhere in the circuit, since the entire circuit is arranged in a series. so $I=I_1=0.1 \text{ A}$.

We now have enough information to find the equivalent resistance of the two resistors, using Ohm's Law – equation (1). We write: $$R_{\text{1 and 2, equivalent}}=\frac{\Delta V_1 + \Delta V_2}{I}=60 \Omega$$ Now, equation (3) tells us $R_{eq}=R_1+R_2$, so $$R_2=R_{eq}-R_1 = 50\Omega$$ The power dissipated across Resistor 2 can be found using the same rewriting of equation (2) as above: $$P_2=I^2R_2= 0.6 \text{ W}$$