Resistors In DC Circuit

We can utilise Kirchhoff’s Law to identify simple rules that apply for resistors in a DC Circuit

In Series §

Take a simple resistor involving a battery and two resistors:

Using the conservation of energy from kirchhoff’s law we get:

$$V_{batt}=\sum IR$$

Because we have no junctions in this circuit, the current, $I$, is constant. So:

$$V_{batt}=I{R}_1+I{R}_2$$

Equivalently, if we define $R_T$ to be an equivalent resistor that could replace $R_1$ and $R_2$, we have:

$V_{batt}=I{R}_T=I(R_1+R_2)$

Therefore, in series (no junctions):

$$R_T=R_1+R_2+R_3+\dots$$

Analogy to geometrical properties

Remember that the Resistance of an object is defined as: $R=\frac{\rho l}{A}$. When we have resistors in series, we are essentially increase the total length, $l$

In Parallel §

In a parallel circuit, we have junctions (marked with dots), where a wire splits paths. Let’s first start by defining the potential through a resistor $R_n$ using Ohm’s Law:

$$V_1=I_1{R}_1{V}_2=I_2{R}_2$$

Recall Kirchhoff’s law for junctions:

$$I_{in}=I_{out}$$

If we identify the splits to have currents $I_1$ and $I_2$ , and the current exiting and entering the battery to be $I$:

$$I=I_1+I_2$$

Now recall we have two possible loops (technically three but we only need two), and that in each loop there is only one component, so:

V = V_{1} = V_{2}$$ Finally, if we define $R$ to be the total resistance:

I_{1} = \dfrac{V_{1}}{R_{1}} \ \ \ \ I_{2} = \dfrac{V_{2}}{R_{2}}

R = \dfrac{V}{I} = \dfrac{V}{\dfrac{V_{1}}{R_{1}} + \dfrac{V_{2}}{R_{2}}}

\dfrac{1}{R} = \dfrac{\dfrac{V}{R_{1}} + \dfrac{V}{R_{2}}}{V}

\dfrac{1}{R} = \dfrac{1}{R_{1}} + \dfrac{1}{R_{2}}

$$whichcanbegeneralisedto:$$

R_{T} = (\dfrac{1}{R_{1}} + \dfrac{1}{R_{2}} + \dfrac{1}{R_{3}} +\dots \dfrac{1}{R_{N}})^{-1}

>[!info] Analogy to geometrical properties > >Remember that the [Resistance](Resistance.md) of an object is defined as: $R = \dfrac{\rho l}{A}$. When we have resistors in parallel, we are essentially increase the total *area*, $A$.