Question

Define Ohm's Law with an example​

Answer 1

Answer:

Ohm’s law states the relationship between electric current and potential difference. The current that flows through most conductors is directly proportional to the voltage applied to it. Georg Simon Ohm, a German physicist was the first to verify Ohm’s law experimentally.

Explanation:

The main application of Ohm's Law is used when building electrical devices. Most appliances need a certain amount of voltage and current to operate. ... Resistors are very important in electric circuits. They are used to limit the amount of current traveling through the circuit and to establish certain levels of voltage.

Answer 2

GiveN :-

• Define Ohm's Law with an example.

SolutioN :-

Ohm's Law establish the relationship between Voltage , resistance and the current. According to Ohm's law if the physical conditions are remained unchanged such as Temperature then the potential difference applied across the ends of the conductor is directly proportional to the current flowing through it . That is ,

$\sf\to \pink{Voltage \propto Current} \qquad \bigg\lgroup \red{\sf Temperature\ constant }\bigg\rgroup$

To remove the proportionality , we use a constant R , which is the resistance of the conductor . So that ,

$\sf\to\pink{ Voltage = (Current )(Resistance)}\qquad \bigg\lgroup \red{\sf Temperature\ constant }\bigg\rgroup$

The conductor which follow Ohm's law are called as Ohmic Conductor and which not are called Non - Ohmic conductor . The V-I graph of a Ohmic conductor is a straight line passing through the origin O . This is not in case of non - ohmic conductors . The V-I graph of both are as follows :-

$\rule{200}2$

G R A P H S :-

$\red{\bigstar} \underline{\textsf{ Graph of Ohmic Conductors :- }}$

$\boxed{\setlength{\unitlength}{1 cm}\begin{picture}(6,6)\thicklines\put(1,1){\vector(1,0){4}}\put(1,1.001){\vector(0,1){4}}\put(1,0.999){\line(1,1){2.5}} \put(0.3,0.7){ \sf O }\put(5.3,1){ \sf I } \put(1,5.3){ \sf V } \put(.8,0.2){ \footnotesize{\sf V-I \ graph \ for \ Ohmic \ Conductor} }\put(3,5){ \boxed{\sf @RishabhRanjan} } \end{picture} }$

$\rule{200}1$

$\red{\bigstar} \underline{\textsf{ Graph of Non- Ohmic Conductors :- }}$

$\boxed{\setlength{\unitlength}{1 cm}\begin{picture}(6,6)\thicklines\put(1,1){\vector(1,0){4}}\put(1,1.001){\vector(0,1){4}} \qbezier(1, .99)( 3, 2)(4,4)\put(0.3,0.7){ \sf O }\put(5.3,1){ \sf I } \put(1,5.3){ \sf V } \put(0,0.2){ \footnotesize{\sf V-I \ graph \ for \ Non-Ohmic \ Conductor} }\put(3,5){ \boxed{\sf @RishabhRanjan} } \end{picture} }$

$\rule{200}2$

S O L V E D E X A M P L E :-

1) $\textsf{ \gray{ The resistances of a conductor is 3 \Omega }}\\\textsf{\gray{ and the applied voltage is 6 Volts . }}\\\textsf{\gray{Find the current flowing through it . }}$

Sol - By Ohm's law we know that V = IR . Using this we have ,

$\sf\dashrightarrow V = I R \\\\\sf \sf\dashrightarrow 3V = I( 3\Omega ) \\\\\sf\dashrightarrow I =\dfrac{3}{3} A \\\\\sf\dashrightarrow\underset{\blue{\sf Required\ Current }}{\underbrace{\boxed{\pink{\frak{ Current = 1\ Ampere }}}}}$