Lesson : Series Circuits – HyperElectronic

In the past lessons, we studied simple circuit with only one resistor directly connected to a direct current (D.C) voltage source. The voltage across the resistor was always the same as the voltage source. This case is not really realistic since we rarely have only one resistor connected to a voltage source. In this lesson, we are going to learn about series circuits. We are going to look at how to find current and voltages in a series circuit.

Series circuits is a circuit with multiples resistors connected together on a single path like in figure 1. In this case, we only have two resistors but you could have a huge amount of resistors in series if you want. The current flowing through the resistors is the same for all resistors in a series circuit. The voltage is divide among the resistors and it is not divided equally (it can be if all resistors have the same value). In the image below, we have a 12V D.C voltage source but the voltage is going to be split between the two resistors. V1 for the voltage across R1 and V2 for the voltage across R2. V1 + V2 needs to be equal to 12V. There is multiple way we could calculated the voltage of each resistors. We will start with the longest method because it help understand how the voltage is divided across the resistors then we will learn the fastest way to calculate the voltage across both resistors.

Figure 1 : Series circuit with two resistors

We can calculate the voltage across both resistor by first calculating the current flowing through the circuit. Once we know the current flowing into the circuit, we can use Ohm’s law to calculate the voltage of each resistors. We have a problem, we can’t calculate the current when we have two resistors. We need to find an equivalent circuit with only one resistor so we can use Ohm’s law. In a series circuit, you only need to add R1 and R2 to get the equivalent circuit.

Figure 2 : Equivalent circuit for a series circuit

We can now calculate the current flowing in the circuit. We will use Ohm’s law for this.

$I = \cfrac{V}{R}=\cfrac{12V}{Req}=\cfrac{12V}{R_1+R_2}$

If R1 = 100 ohms and R2 = 50ohms, the current would be equal to:

$I = \cfrac{12V}{100\Omega+50\Omega}=\cfrac{12V}{150\Omega}=0.08A=80mA$

We can now calculate the voltage across both resistors using Ohm’s law:

$V = R\cdot I$

$V_1 = R_1\cdot I=100\Omega \cdot 0.08A = 8V$

$V_2 = R_2\cdot I=50\Omega \cdot 0.08A = 4V$

We could have calculate only V1 or V2 since we know that V1 + V2 = 12V.

$V_{source} = V_1 + V_2 = 12V$

$12V - V_1 = V_2$

$V_2 = 12V - 8V = 4V$

$12V - V_2 = V_1$

$V_1 = 12V - 4V = 8V$

We can see here that the higher resistance will get a bigger voltage difference. Calculating the voltages this way is pretty long but it was done to show how to calculate the equivalent circuit of resistors in series. We will now look at the fastest way to calculate the voltage across a resistors when they are in series. This technique is a bit more complex to understand and will be seen again in the voltage divider lesson.

The equivalent circuit we found earlier will be useful here. We know that:

$1) V_{source} = 12V = Req \cdot I = (R_1+R_2) \cdot I$

and

$2) V_1 = R_1 \cdot I$

$3) V_2 = R_2 \cdot I$

We can isolate I in the first formula and replace it in the equation 2) and 3):

$I=\cfrac{V}{(R_1+R_2)}=\cfrac{12V}{(R_1+R_2)}$

$V_1=R_1\cdot I=\cfrac{12V\cdot R1}{(R1+R2)}=\cfrac{12V\cdot 100\Omega}{(100\Omega+50\Omega)}=8V$

$V_2=R_2\cdot I=\cfrac{12V\cdot R2}{(R1+R2)}=\cfrac{12V\cdot 50\Omega}{(100\Omega+50\Omega)}=4V$

We this technique, we don’t need to calculate the current flowing into the circuit which reduce the number of equation to solve. In most real life application, you will know the value of the resistors and the voltage so you can easily use this formula to find out the voltage on each resistors. Even if you have more than two resistors, you can use this. The generic formula is:

$V_x=V\cdot \cfrac{R_x}{R_x+R_n+R_{n+1} +R_{n+2} +\dots}$

where Vx is the voltage across the resistor Rx and V is the voltage across all resistors in series.