Question

When a 4 ohm resistor is connected across the terminals of a2V battery, the number of coulombs passing through the resistor per second is 0.5 1 2 4

Answer 1

Given :

▪ An identical resistor of resistance 4Ω is connected with a battery of 2V.

To FinD :

▪ No. of charges passing through the resistor per second.

Concept :

☞ This question is completely based on the concept of Ohm's law.

☞ As per this law, current passing through a conductor is directly proportional to the applied potential difference. (At constant temp.)

Mathematically,

$\bigstar\:\underline{\boxed{\bf{\red{V=I\times R}}}}$

• V denotes applied voltage
• I denotes current flow
• R denotes resistance

Calculation :

$\dashrightarrow\sf\:V=I\times R\\ \\ \dashrightarrow\sf\:V=\dfrac{Q}{t}\times R\\ \\ \dashrightarrow\sf\:Q=\dfrac{V\times t}{R}\\ \\ \dashrightarrow\sf\:Q=\dfrac{2\times 1}{4}\\ \\ \dashrightarrow\underline{\boxed{\bf{\blue{Q=0.5C}}}}\:\gray{\bigstar}$

Answer 2

Given that ,

" When a 4 Ω resistor is connected across the terminals of a 2 V battery "

We need to find the number of columbs i.e., charges passing through the resistance per second . This is nothing but charge

Resistance , R = 4 Ω

Voltage , V = 2 V

Time , t = 1 s

Current , I = ? A or ? Columb / second

Now use Ω 's law to find the current in the circuit .

⇒ V = IR

⇒ 2 = I ( 4 )

⇒ I = 0.5 Amperes

⇒ I = 0.5

Now Current is defined as charge per second

$\implies \bf I=\dfrac{Q}{t}\\\\\implies \bf Q=It\\\\\implies \bf Q=0.5*1\\\\\implies \bf Q=0.5\ Columb$