3.2 Voltage and Current Dividers

Voltage Dividers. Consider the circuit shown in figure 3.10 having three identical resistors connected in series to a 6 V battery. What is the voltage drop across each resistor? Intuition might correctly lead the reader to the answer of 2 V: the 6 V battery is connected in series with three identical resistors.  The battery’s 6 V must be dropped across the series set of resistors; the resistors are identical; there should be \frac{1}{3} of 6 V or 2 V across each resistor. We can readily obtain this result by applying KVL around the circuit loop and making use of the fact that the current is the same for all elements in the series-connected circuit, hence the voltage drop across each resistor is the same. Application of KVL around the loop gives:

(1)   \begin{equation*}  -6+V_{R} + V_{R} + V_{R} = 0 \end{equation*}

which readily gives us V_{R} = 2 V.

Figure 3.10 6V battery connected in series to three identical resistances. The voltage drop across each resistor is 2V

This is a voltage divider circuit. Recall from the previous chapter that a series-connection of three resistors  each having resistance R\Omega is equivalent to a single resistor having resistance 3R\Omega. In this circuit the resistance seen by the battery, that is, the equivalent resistance loading the battery, is 3R\Omega.

Knowing this, we deduce that the current in the loop is

(2)   \begin{equation*}  I=\frac{6}{3R}=\frac{2}{R} \end{equation*}

Having found the current, we can determine the voltage drop across each resistor via Ohm’s law:

(3)   \begin{equation*}  V_{R}=IR=\frac{2}{R}R=2 \end{equation*}

Note that the resistance of each resistor is 1/3 of the total series resistance seen by the battery, and the voltage drop across each resistor is 1/3 of the battery voltage. This is the behavior of a voltage divider circuit which we will define below.  We first extend the discussion to the case where the series-connected resistors have different values.

Consider the following network which has three series resistors each having a different value, and let us determine the voltages across R_{1}, R_{2}, and R_{3}.

Figure 3.11

To begin, we note that the three resistors are all connected in series; this circuit is equivalent to the following circuit, in which the voltage source is applied across a single resistor having equivalent resistance

(4)   \begin{equation*}  R_{eq}=R_{1}+R_{2}+R_{3} \end{equation*}

Figure 3.12 Circuit of figure 3.11 with the set of series resistors replaced by a single equivalent resistance

The current in the loop is

(5)   \begin{equation*}  i=\frac{v_{s}}{R_{eq}}=\frac{v_{s}}{R_{1}+R_{2}+R_{3}} \end{equation*}

The current having been determined, we readily obtain the voltage across each of the resistors via Ohm’s law:

(6)   \begin{equation*}  v_{R1} = iR_{1} = v_{s} \frac{R_{1}}{R_{eq}} =v_{s} \frac{R_{1}}{R_{1}+R_{2}+R_{3}}\end{equation*}

(7)   \begin{equation*}  v_{R2} = iR_{2} = v_{s} \frac{R_{2}}{R_{eq}}=v_{s} \frac{R_{2}}{R_{1}+R_{2}+R_{3}} \end{equation*}

(8)   \begin{equation*}  v_{R3} = iR_{3} = v_{s} \frac{R_{3}}{R_{eq}} = v_{s} \frac{R_{3}}{R_{1}+R_{2}+R_{3}} \end{equation*}

We thus  see that the voltage drop across each series resistor is a fraction of the source voltage; the fraction is the ratio of the series resistance to the total resistance seen by the voltage source.

Note that the sum of equations (6), (7), and (8) gives:

(9)   \begin{equation*}  v_{R1} + v_{R2} + v_{R3} = v_{s}\frac{R_{1}+R_{2}+R_{3}}{R_{eq}} = v_{s}\frac{R_{1}+R_{2}+R_{3}}{R_{1}+R_{2}+R_{3}} = v_{s} \end{equation*}



Voltage Divider Example. Determine the voltages across the five resistors in the following circuit if R_{1}=100\Omega, R_{2}=200\Omega, and R_{3}, R_{4} and R_{5} are all 1k\Omega resistors.

Figure 3.13a

Solution: This problem can be solved by combining resistances as shown in Figure 3.13b, then forming a single equivalent resistance, determining the current from the 6V battery, and applying KVL, KCL, and Ohm’s law to arrive at the results. Using voltage division arrives at the result more quickly when we realize that there is a 6V drop across the middle of the circuit (R{_1}+R_{2}) and the right branch of the circuit (R_{3}+R_{4}+R_{5}). Voltage division gives the voltage across R_{1} as 6\frac{R_{1}}{R_{1}+R_{2}}=6\frac{100}{300}=2V. Similarly, the voltages across R_{2}, R_{3}, R_{4}, and R_{5} are 4V, 2V, 2V, 2V.

Figure 3.13b


Example: A voltage divider is used to to create a 5V voltage supply from a 9V battery by use of  400\Omega and 500\Omega resistors as shown.  Determine the output voltage when a 10k\Omega load is attached to the output. Repeat for a 1k\Omega, 100\Omega, and 10\Omega loads.


Figure 3.14a Unloaded voltage divider circuit

Solution: Without a load resistance, the output voltage from this voltage divider is 9\frac{500}{400+500}=5V. When the load resistor is added to the circuit as shown in Figure 3.14b, the voltage division changes. The 9V battery voltage divides into a voltage across the 400\Omega resistor and the parallel combination of the 500\Omega resistor and the added load resistance, R_{L}. Since this parallel combination will be less than 500\Omega, the output voltage will be less than in the non-loaded case.

A 10k\Omega load in parallel with 500\Omega is 476\Omega. The output voltage in this case will be 9\frac{476}{400+476}=4.89V

A 1k\Omega load in parallel with 500\Omega is 333\Omega. The output voltage in this case will be 9\frac{333}{400+333}=4.08V

A 100\Omega load in parallel with 500\Omega is 83.3\Omega. The output voltage in this case will be 9\frac{83.3}{400+83.3}=1.55V

A 10\Omega load in parallel with 500\Omega is 9.8\Omega. The output voltage in this case will be 9\frac{9.8}{400+9.8}=0.21V

Clearly, the effect of “loading down” the output by the addition of a parallel load resistance is to reduce the output voltage; decreasingly smaller load resistances represent increased loading, in the sense that smaller and smaller load resistances produce greater changes in the behavior of the voltage divider circuit. An infinite load resistance, R_{L}=\infty would cause no change in the voltage divider, and we would have V_{o}=5V. A 0\Omega load resistance would result in V_{o}=0V.

Figure 3.14b Voltage divider circuit loaded down with load resistance R_{L}


Example:  Considering the behavior of the loaded-down voltage divider of the previous example,  design a voltage divider circuit to provide an unloaded 4.5V output from a 9V input. When loaded with resistance values between 100kΩ and 100Ω, the output voltage should not drop below 10% of the desired 4.5V value. 

Figure 3.15a

Solution: With reference to figure 3.15a, in the unloaded case, any equal value R_{1}=R_{2}=R will divide 9V into the required 4.5V output voltage. We know from the previous example problem that load resistances will cause V_{o} to decrease below 4.5V, and we know that smaller load resistances (which, in effect, represent “larger loads” owing to their larger impact on the circuit) will cause larger decreases than larger load resistances. The problem specifies that V_{o} should not drop below 4.5V-10%, or 4.05V so R should be chosen to provide a 4.05V output voltage when the circuit is loaded down with a 100\Omega resistor. This problem thus is reduced to determining the value R for the circuit shown in figure 3.15b.  A value R=22.2\Omega will result in the desired output voltage when the circuit is loaded with a 100\Omega resistance. The algebra is left as an exercise for the reader to show.

Figure 3.15b


Current Dividers. Consider the circuit shown, where a current i is divided between two circuit branches, with branch resistances R_{1} and R_{2}.

Figure 3.16 Current divider circuit. Current I divides into two branches, i_{1} and i_{2}

We wish to determine how current i divides or separates, into currents i_{1} and i_{2}.  We begin by noting that, via Ohm’s law,

(10)   \begin{equation*}  i_{1}=\frac{v}{R_{1}} \end{equation*}


(11)   \begin{equation*}  i_{2}=\frac{v}{R_{2}} \end{equation*}

Since we are seeking an expression giving i_{1} and i_{2} in terms of i we need an expression relating v and i.  We replace the two parallel resistors by their equivalent resistance, 

(12)   \begin{equation*}  R_{eq}=\frac{R_{1}R_{2}}{R_{1}+R_{2}} \end{equation*}

from which we obtain

(13)   \begin{equation*}  v=iR_{eq}=i\frac{R_{1}R_{2}}{R_{1}+R_{2}} \end{equation*}

Substituting eq (13) into eq (10) and (11) we find

(14)   \begin{equation*}  i_{1}=i\frac{R_{eq}}{R_{1}}=i\frac{R_{2}}{R_{1}+R_{2}} \end{equation*}


(15)   \begin{equation*}  i_{2}=i\frac{R_{eq}}{R_{2}}=i\frac{R_{1}}{R_{1}+R_{2}} \end{equation*}

This result shows how the current divides fractionally into the two parallel paths. We thus  see that the current through each parallel resistor is a fraction of the entering current; the fraction for one resistor is the ratio of the resistance of the other resistor to the total resistance.


Example: In the following circuit, R_{1}=1\Omega and R_{2}=1k\Omega. If a current I=1A flows into this network, determine the current flowing through R_{1} and R_{2}

Figure 3.15

Solution:  I_{1}=I \frac{R_{2}}{R_{1}+R_{2}}=1 \frac{1000}{1001}=0.999A while I_{2}=I \frac{R_{1}}{R_{1}+R_{2}}=1 \frac{1}{1001}=0.001A  This example well illustrates the common maxim “current takes the path of least resistance”.




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Applied Electrical Engineering Fundamentals by David J. McLaughlin is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

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