Difference between revisions of "Voltage and Current Dividers"

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==Current Division==
==Current Division==
Like the voltage divider, resistors in parallel will divide up the current. When we have a current flowing through resistors in parallel, we can express the current flowing through a single resistor as ratio of currents and resistances, without ever knowing the voltage.
Resistors in parallel divide up the current. When we have a current flowing through resistors in parallel, we can express the current flowing through a single resistor as ratio of currents and resistances, without ever knowing the voltage.


[[Image:current_division1.gif]]
[[Image:current_division1.gif]]

Revision as of 15:34, 5 July 2006

Voltage Division

When we have a voltage across a string of resistors connected in series, the voltage across the entire string will be divided up among the resistors. We can express the voltage across a single resistor as a ratio of voltages and resistances, without ever knowing the current.

Voltage division1.gif

In the circuit above,

or

We can generalize this equation for number of resistors in series with the equation:

where is the voltage across resistor </math>k is the voltage across the whole string of resistors.

Current Division

Resistors in parallel divide up the current. When we have a current flowing through resistors in parallel, we can express the current flowing through a single resistor as ratio of currents and resistances, without ever knowing the voltage.

Current division1.gif

In the circuit above

or

where is the current flowing through all the resistors. Note that the numerator on the right is R2, not R1. Remember that a larger resistance will carry a smaller current.

We can generalize the equation for number of resistors in parallel with the equation:

where is the current flowing through resistor and is the current flowing through all the resistors.

Practice Problems

Problem 1

Use voltage division to find in the circuit below:

Voltage division problem1.gif

click here for the solution

Problem 2

Simplify the circuit and then use current division to find in the circuit below:

Current division problem1.gif

click here for the solution