Voltage Division in a Series Circuit

The applied source voltage is divided into voltage drops through all resistive elements in the series circuit. So a question arises here that,
How will a resistor's value affect the voltage across the resistor?
It can be sensed and have tested that the value of voltage drops depends upon the resister value in the circuit.
It turns out that in a series circuit, the larger the resistance the larger the voltage across the resistance.
In fact, there is a ratio rule that states that
The ratio of the voltages across series resistors is in direct proportion to the ratio of their resistive values.
$$ \bbox[5px,border:1px solid red] {\color{blue}{\frac{V_1}{V_2} = \frac{R_1}{R_2}}} \tag{1}$$
How Eq. (1) is calculated, lets practice on it.
The previous sections demonstrated that the current remains the same for all the elements in the series circuits. The voltage drops for different resisters can be written as
$$V_1 = IR_1$$
$$V_2 = IR_2$$
In the above two equations, current I is same for both the equations.
$$ I = \frac{V_1}{R_1} $$
$$ I = \frac{V_2}{R_2} $$
Equalizing both the above equations, we can get
$$\frac{V_1}{R_1}= \frac{V_2}{R_2} $$
$$\frac{V_1}{V_2}= \frac{R_1}{R_2} $$
For any resister and its voltage drop combination,
Ratio Rule:
$$ \bbox[5px,border:1px solid blue] {\color{blue}{\frac{V_i}{V_j} = \frac{R_i}{R_j}}} \tag{2}$$
Example 1: Using the information provided in [Fig. 1], find
a. The voltage V1 using the ratio rule.
b. The voltage V3 using the ratio rule.
c. The applied voltage E using Kirchhoff's voltage law.
Fig. 1: Circuit to be examined in example 1.
a. Applying the ratio rule:
$$\frac{V_1}{V_2}= \frac{R_1}{R_2} $$
$$\frac{V_1}{6V}= \frac{6 Ω}{3 Ω} $$
$$V_1= 2 (6V) = 12V$$
b. Applying the ratio rule:
$$\frac{V_2}{V_3}= \frac{R_2}{R_3} $$
$$\frac{6V}{V_3}= \frac{3 Ω}{1 Ω} $$
$$V_3= \frac{1}{3} (6V) = 2V$$
c. Applying Kirchhoff's voltage law:
$$ E = V_1 + V_2 + V_3$$
$$E = 12V + 6V +2V = 20V $$

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