#### Recommended Level

Beginner

### Kirchoff's Voltage Law

Kirchhoff's Voltage Law states that that the algebraic sum of all the voltages in a loop must equal zero. A practical application of this law is the voltage divider, shown in the figure below.

If U_{1} is a 9-V battery, R_{1} a 3-Ohm resistor, and R_{2} a 6-Ohm resistor, the current flowing in the circuit is

$$\frac{9 \text{ v}}{3 \text{ } \Omega + 6 \text{ } \Omega} = 1 \text{ A}$$.

The voltage across R_{1} would then be

$$ 3 \text{ }\Omega \text{ x } 1 \text{ A} = 3\text{ v}$$,

and the voltage across R_{2} would be

$$ 6 \text{ }\Omega \text{ x } 1 \text{ A}= 6\text{ v}$$.

The voltage across the individual resistors is equal to the voltage across both resistors times the ratio of the individual resistance to the total resistance. The equation for calculating the voltage across resistor R_{2}, U_{2}, is also shown below.

Fig. 1 Voltage Divider Circuit

#### $$U_{1}=\frac {U_{1}R_{2}}{R_{1}+R_{2}}$$

### Kirchoff's Current Law

Similarly, you can divide currents. The law that governs this is called Kirchoff's Current Law. This law states, "The algebraic sum of all currents entering and exiting a node must equal zero."

To see how this works, refer to the figure below. I_{T} is the total current being supplied to the circuit of four parallel resistors. I_{X} is the current that will flow through R_{X}. As shown by the equation below, the fraction of I_{T} flowing through R_{X }will be equal to I_{X} times the ratio of R_{X} to the total resistance. If all of the resistors are the same value, then the current through each will be the same.

For example, if $$I_{T} = 1 \text{ A}$$ and $$R_{X}=R_{1}=R_{2}=R_{3}=30 \;\Omega$$,

then $$R_{T} = 10 \;\Omega$$ and $$I_{X} = \frac{I_{T}R_{T}}{R_{X}+R_{T}} = \frac{(1)(10)}{30+10} = \frac{10}{40} = 0.25 \text{ A}$$.

Fig. 2 Current Divider Circuit

#### $$I_{X}=\frac{I_{T}R_{T}}{R_{X}+R_{T}}$$

### Applications of Voltage and Current Dividers

Voltage and current dividers have many practical applications. For example, in the circuit shown below, R_{2 }is a photoresistor, which changes value as the intensity of the light impinging on it changes. The voltage across R_{2}, U_{2}, is then a measure of light intensity. You could use this to sense the level of sunlight, or the presence or absence of an object between a light source and the photoresistor. There are, of course, many other applications for voltage and current divider circuits.

Fig. 3 Voltage Divider Application

#### $$U_{2}=\frac{U_{1}R_{2}}{R_{1}+R_{2}}$$

1 CommentLoginPete Dubler2017-07-27This article would be easier to understand if U2 were illustrated as a volt meter instead of as a battery. U2 is not a voltage “source” like a battery. This confuses things. One measures the voltage here across R2. There is no voltage source applied across R2 by U2. Also, in Fig. 1, the equation should read “U2 =...”, not “U1 =...”