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A series connected circuit is often referred to as a *voltage divider circuit. *The source voltage equals the total of all voltage drops across the series connected resistors. The voltage dropped across each resistor is proportional to the resistance value of that resistor. Larger resistors experience larger drops, while smaller resistors experience smaller drops. The

where *V _{X}* = voltage dropped across selected resistor

*R _{X}* = selected resistorâ€™s value

*R _{T} = *total series circuit resistance

* V _{S}* = source or applied voltage

A simple example to start:

Example 1

Find the voltage drop across each resistor, given that V=150 V, R = 1 Kohm.

The first solution requires that we find the series current. First, calculate the circuitâ€™s total resistance: R_{tot} = R_{1} + R_{2} = 1k+2k = 3 kohm.

Next, find the circuit current: I = V/R_{tot} = 150/3 = 50 mA.

Finally, find the voltage across R_{1}: V_{1}= I R_{1} = 50 V;

and the voltage across R_{2}: V_{2} = I R_{2} = 100 V.

The second, more direct solution uses the voltage divider formula:

and

{Solution by TINA's Interpreter!}

I:=V/(R+2*R);

VR:=I*R;

V2R:=I*2*R;

VR=[50]

V2R=[100]

{or using the voltage divider formula:}

VR:=V*R/(R+2*R);

V2R:=V*2*R/(R+2*R);

VR=[50]

V2R=[100]

Another example:

Example 2

Find the voltage drop on each resistors.

Use the voltage divider formula:

{Solution by TINA's Interpreter!}

{Use the voltage divider formula: Vi= Vs*Ri/Rtot}

V1:=VS*R1/(R1+R2+R3+R4);

V2:=VS*R2/(R1+R2+R3+R4);

V3:=VS*R3/(R1+R2+R3+R4);

V4:=VS*R4/(R1+R2+R3+R4);

V1=[500m]

V2=[1]

V3=[1.5]

V4=[2]

Example 3

Find the voltages measured by the instruments.

This example shows that the branch connected in parallel with the source does not affect the use of the voltage division formula.

{Solution by TINA's Interpreter}

V1:=V*R3/(R3+R4);

V1=[100]

V2:=V*R4/(R3+R4);

V2=[100]

The following example is a bit more complicated:

Example 4

Find the voltage drop across R_{2} if the voltage source is 140 V and the resistances are as given in the schematic.

{Solution by TINA's Interpreter!}

V4:=Vs*(Replus(R4,(R2+R3)))/(R1+Replus((R2+R3),R4));

V:=V4*R2/(R2+R3)

{or}

Sys I,I2,I1,V

I*R4=I2*(R2+R3)

I1=I+I2

V=I2*R2

Vs=R1*I1+I*R4

end;

V=[40]

The voltage division formula is used twice, first to find the voltage across R4, and second to find the voltage across R2.

Example 5

Find the voltage between the nodes A and B.

Use the voltage division formula three times:

The method here is to first find the voltage between the ground node and the node (2) where R2, R3, and R1 are joined. This is done using the voltage divider formula to find the portion of Vs appearing between these two nodes. Then the voltage divider formula is used twice to find Va and Vb. Finally, Vb is subtracted from Va.

{Solution by TINA' Interpreter!}

R12:=Replus((R1+R2),(R1+R2+R3));

V12:=Vs*R12/(R2+R12);

Vab:=V12*(R2/(R1+R2)-R1/(R1+R2+R3));

Vab=[500m]