Question

Explain ohm's law .........

Answer 1

Answer:

Ohm's Law states that electric current flowing in a conductor is directly proportional to potential difference across the ends of the conductor,provided temperature and other physical conditions remains constant.

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Answer 2

Hi mate

Here is ur answer

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Ohm’s Law states that the current flowing through a conductor is directly proportional to the potential difference applied across its ends, provided the temperature and other physical conditions remain unchanged. Mathematically it can be represented as,

Potential difference ∝ Current is

$V ∝ I$

( When the value of V increases the value of I increases simultaneously)

V = IR

Where,

• V is Voltage in volts (V)
• R is Resistance in ohm (Ω)
• I is Current in Ampere (A)

How does Ohm’s Law work?

Ohm’s Law describes the current flow through a resistance when different electric potentials (voltage) are applied at each end of the resistance. Since we can’t see electrons, the model or an analogy of electric circuits used to help us understand circuits better is the water-pipe analogy. Water flowing through pipes is a good mechanical system that is analogous to an electrical circuit.

Here, the voltage is analogous to water pressure, the current is the amount of water flowing through the pipe, and the resistance is the size of the pipe. More water will flow through the pipe (current) when more pressure is applied (voltage) and the bigger the pipe, (lower the resistance).

Different Applications of Ohm’s Law

The main applications of Ohm’s law are:

• To determine the voltage, resistance or current of an electric circuit.
• Ohm’s law is used to maintain the desired voltage drop across the electronic components.
• Ohm’s law is also used in dc ammeter and other dc shunts to divert the current.

Calculating Different Parameters Using Ohm’s Law

As an equation, the Ohm’s Law serves as an algebraic recipe for calculating the current when the resistance and the potential difference are known. Likewise, if any two parameters in the equation are known, then the unknown third parameter can be easily calculated as follows:

To find Voltage(V),

$V = IR$

To find Current(I),

$I = \frac{V}{R}$

To find Resistance(R),

$R = \frac{v}{i}$