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A *voltage divider* or *potential divider* is a series circuit that is used to provide more than one reduced voltages from a single source of voltage.

Consider a circuit of voltage divider as shown below, in which two reduced voltages V_{1} and V_{2} are obtained from a single input voltage source of V volts. Since no load is connected to circuit, it is called *unloaded voltage divider*.

Refer the circuit of unloaded voltage divider,

$$\mathrm{Circuit\:Current,I= \frac{V}{R_{1}+{R_{2}}}=\frac{V}{R_{eq}}}\:\:\:… (1)$$

Where,R_{eq}=R_{1} + R_{2}= Total resistance of voltage divider

Therefore,

$$\mathrm{V_{1}=IR_{1}=\frac{V}{R_{eq}}×R_{1}=V\frac{R_{1}}{R_{eq}}}\:\:\:… (2)$$

$$\mathrm{V_{2}=IR_{2}=\frac{V}{R_{eq}}×R_{2}=V\frac{R_{2}}{R_{eq}}}\:\:\:… (3)$$

Hence, equation (2) and (3) shows that, the voltage drop across any resistor in an unloaded voltage divider is equal to the total source voltage multiplied by the ratio of that resistance value to the total resistance.

When a load resistance R_{L} is connected across the output terminals of the voltage divider, the voltage divider is said to be loaded.

Due to this R_{L}, the output voltage (V_{2} in this case) is reduced by an amount depending on the value of R_{L}. It is because the load resistance R_{L} is in parallel with R_{2} and reduces the effective resistance between the load terminals (here, A and B), as a result the output voltage is reduced. The loading of voltage divider has following effects −

The output voltage decreases depending upon the value of load resistor R

_{L}.After connected the load resistor, the voltage divider circuit turned into a series-parallel circuit. Therefore, the total resistance of the circuit is reduced.

The circuit current increases because the total resistance of the circuit is decreased.

In the voltage divider circuit shown below, determine the following −

- Total current drawn from the supply
- Voltage across the load resistor
- The current flowing through R
_{L} - The current in the tapped portion.

**Solution**

The equivalent resistance of the given voltage divider is,

$$\mathrm{R_{BC\:eq}=(R_{BC}\:||\:R_{L})=\frac{150 × 250}{150 + 250}= 93.75 Ω}$$

$$\mathrm{∴R_{eq}=R_{AB}\:+\:R_{BC\:eq}= 100 + 93.75 = 193.75 Ω}$$

Total current drawn from the source

$$\mathrm{I=\frac{V}{R_{eq}}=\frac{200}{193.75}= 1.032 A}$$

Voltage across load resistor

$$\mathrm{V_{L}=V×\frac{R_{BC\:eq}}{R_{eq}}= 200 ×\frac{93.75}{193.75}= 96.77 Ω}$$

The current flowing through R

_{L}

$$\mathrm{I_{L}=\frac{V_{L}}{R_{L}}=\frac{96.77}{250}= 0.387 A}$$

Current in the tapped portion

$$\mathrm{I_{Bc}=I-I_{L}=1.032 − 0.387 = 0.645 A}$$

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