# Voltage Divider Rule (VDR) – Solved Examples for R, L and C Circuits

## Voltage Division “VDR” for for Resistive, Inductive and Capacitive Circuits

**What is Voltage Divider Rule?**

In a circuit, when a number of elements are connected in series, input voltage divides across the elements. And in a circuit, when a number of elements are connected in parallel, the current divides across the elements.

Therefore, in a parallel circuit, the current divider rule is used and in a series circuit, the voltage divider rule is used to analyze and solve the circuit.

When two or more impedances are connected in series, the input voltage is divided into all impedances. To calculate the voltage across each element, the voltage divider rule is used. The voltage divider rule is the most important and simple rule in circuit analysis to calculate the individual voltage of any elements.

The voltage divider rule is also known as the potential divider rule. In some conditions, we require specific output voltage. But we don’t have that specific value of the source. In this condition, we make a series of passive elements and reduce the voltage level to a specific value. And here, the voltage divider rule is used to calculate the specific output voltage.

According to elements used in the circuit, the voltage divider rule can be classified into three types; resistive voltage divider, inductive voltage divider, and capacitive voltage divider. Now, we will prove the voltage divider rule for all these types of circuits.

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**Voltage Divider Rule for Resistive Circuits**

To understand the resistive voltage divider rule, we take a circuit having two resistors are connected in series with the voltage source.

As the resistors are connected in series, the current that passes through both resistors is the same. But the voltage is not the same for both resisters. The input voltage of the circuit divides into both resisters. And the value of individual voltage depends on the resistance.

As shown in the above figure, two resistors R_{1} and R_{2} are connected in series with the voltage source V_{s}. The total current supplied by the source is I ampere. As all elements are connected in series, it will make a single loop and the current that passes through all elements is the same (I amp).

The voltage across resistor R_{1} is V_{R1} and the voltage across resistor R_{2} is V_{R2}. And the total supplied voltage divides between both resisters. Hence, the total voltage is a sum of V_{R1} and V_{R2}.

*V _{S}* =

*V*

_{R1}+

*V*

_{R2}… (1)

According to Ohm’s law,

*V _{R1}* =

*IR*

_{1}+

*IR*

_{2}… (2)

Therefore, from the equation-(1) and (2);

*V _{S}* =

*IR*

_{1}+

*IR*

_{2}

*V _{S}* =

*I(R*

_{1}+

*R*

_{2})

Now, put the value of current I in the equation-(2);

*V _{R1}* =

*IR*

_{1}

Similarly;

*V _{R2}* =

*IR*

_{2}

Hence, the voltage divider rule for a resistive circuit is opposite to the current divider rule. Here, the voltage of the resistor is a ratio of multiplication of total voltage and that resistance to the total resistance.

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#### Solved Example of Resistive Circuit using VDR

**Example-1 **

Find the voltage across each resistor using the voltage divider rule.

Here, three resistors (R_{1}, R_{2}, and R_{3}) are connected in series with 100V source voltage. The voltage across resistors R_{1}, R_{2}, and R_{3} are V_{R1}, V_{R2}, and V_{R3} respectively.

The voltage across resistor R_{1};

*V _{R}*

_{3}= 500 / 30

*V _{R}*

_{3}= 16.67

*V*

The voltage across resistor R_{2};

*V _{R}*

_{3}= 100 / 30

*V _{R}*

_{3}= 33.33

*V*

The voltage across resistor R_{3};

*V _{R}*

_{3}= 1500 / 30

*V _{R}*

_{3}= 50

*V*

Total voltage V_{T};

*V _{T} =V*

_{R1}+

*V*

_{R2}+

*V*

_{R3}

*V _{T}* = 16.67 + 33.33 + 50

*V*_{T} = 100 V

*V _{T} = V_{S}*

Hence, it is proved that the total voltage= is similar to the supplied voltage.

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**Voltage Divider Rule for Inductive Circuits**

When a circuit having more than two inductors is connected in series, the current that passes through the inductors is the same. But the source voltage is divided into all inductors. In this condition, the voltage across the individual inductor can be found by the inductor voltage divider rule.

Consider as shown in the above figure, two inductors (L_{1} and L_{2}) are connected in series. And total current I pass through the inductor. The voltage across inductor L_{1} is V_{L1} and the voltage across inductor L_{2} is V_{L2}. And the supply voltage is V_{S}. Now, we need to find the voltage V_{L1} and V_{L2} using the inductor voltage divider rule.

As we know the equation of voltage for inductor;

Where L_{eq} is the total inductance of the circuit. Here, two inductors are connected in series. Therefore, the equivalent inductance is a sum of both inductances.

*L _{eq}* =

*L*

_{1}+

*L*

_{2}

From, equation-(3);

Now, the voltage across inductor L_{1} is;

Similarly, the voltage across inductor L_{2} is;

So, we can say that the voltage divider rule for an inductor is the same as the resistors.

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#### Solved Example of Inductive Circuit using VDR

**Example-2**

Find the voltage across each inductor for the given circuit using the voltage divider rule.

Here, two inductors are connected in series with 100V, 60Hz source. The voltage across inductor L_{1} is V_{L1} and the voltage across inductor L_{2} is V_{L2}.

To find the voltage across inductors, we need to find the reactive impedance of each inductor.

The reactive impedance across inductor L_{1} is;

X_{L1} = 2 π *f *L_{1}

X_{L1} = 2 × 3.1415 × 60 x 10 × 10^{-3}

X_{L1} = 3.769 Ω

The reactive impedance across inductor L_{2} is;

X_{L2} = 2 π *f *L_{2}

X_{L2} = 2 × 3.1415 × 60 x 14 × 10^{-3}

X_{L2} = 5.277 Ω

According to the voltage divider rule,

The voltage across inductor L_{1} is;

*V*_{L1} = 41.66 *V*

The voltage across inductor L_{2} is;

V_{L2} = 58.35 *V*

Total voltage V_{T} is;

*V _{T} =V*

_{L1}+

*V*

_{L2}

*V _{T}* = 41.66 + 58.35

*V*_{T} = 100 *V*

*V _{T} = V_{S}*

So, the total voltage is the same as supplied voltage.

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**Voltage Divider Rule for ****Capacitive**** ****Circuits**

In a capacitor, the voltage divider rule is different compare to the inductor and resistor. To calculate the voltage divider rule for capacitors, let’s consider a circuit having two or more capacitors connected in series.

Here, two capacitors are connected in series with the source voltage V_{S}. The source voltage divides into two voltages; one voltage is across the capacitor C_{1} and a second voltage is across the capacitor C_{2}.

The voltage across capacitor C_{1} is V_{C1} and the voltage across capacitor C_{2} is V_{C2}. As shown in the above circuit diagram, both capacitors are connected in series. Therefore, the equivalent capacitance is;

The total charge delivered by the source is Q;

*Q* = *C*_{eq} V_{S}

The voltage across capacitor C_{1} is;

*V _{C1}* =

*Q*

_{1}/

*C*

_{1}

The voltage across capacitor C_{2} is;

*V _{C2}* =

*Q*

_{1}/

*C*

_{2}

So, from the calculation, we can say that the individual voltage across the capacitor is a ratio of multiplication of total source voltage and opposite capacitance to the total capacitance.

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#### Solved Example of Capacitive Circuit using VDR

**Example-3**

Find the voltage across each capacitor for a given network using the voltage divider rule.

Here, two capacitors are connected in series with 100 V, 60 Hz source. The voltage across capacitor C_{1} is V_{C1} and the voltage across capacitor C_{2} is V_{C2}.

To calculate the voltage across each capacitor, we need to find the capacitive impedance.

The capacitive impedance across C_{1} is;

The capacitive impedance across C_{2} is;

*X*_{C2} = 1 / (2 π *f *C_{2})

*X*_{C2} = 1 / (2 π × 60 × 20 ×10^{-6})

*X*_{C2} = 10^{-6} / 7539.822

*X*_{C2} = 132.63 Ω

According to the voltage divider rule, the voltage across capacitor C_{1} is;

*V*_{C1} = 33.33 *V*

the voltage across capacitor C_{2} is;

*V*_{C2} = 66.67 *V*

The total voltage across capacitor V_{T} is;

*V _{T} =V*

_{C1}+

*V*

_{C2}

*V _{T}* = 33.33 + 66.67

*V*_{T} = 100 *V*

*V _{T} = V_{S}*

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