Question

three resistor 2 ohm 3 ohm and 4 ohm is connected in series this combination is connected to a source of 10 volt what is the potential drop across each resistor​

Answer 1

Answer :-

Given :-

• Value of resistors = 2 ohm , 3 ohm & 4 ohm

• Potential difference (V)= 10 volt

• Type of Combination = Series

Now as the resistors are connected in series

R equivalent = 2 + 3 + 4

= 9 ohm

Now as IR = V

=> I × 9 = 10

=> I = 10 ÷ 9

=> I = 1.11...

Or I = 1.1

Now As as potential drop V' = IR'

For 2 ohm resistor

V' = 1.1 × 2

V' = 2.2 volt

For 3 ohm resistor

V' = 1.1 × 3

V' = 3.3 volt

For 4 ohm resistor

V' = 1.1 × 4

V' = 4.4 volt

Answer 2

$\huge\bold{\underline{\underline{Answer:-}}}$

Let R1, R2 and R3 be the resistors of 2 ohm, 3 ohm and 4 ohm respectively.

We know that in series combination, Equivalent resistance R-

R = R1 + R2 + R3

$\large\sf{\implies R = (2 + 3 +4) \:ohm}$

$\large\sf{\implies R = 9 \:ohm}$

Given that potential difference, V = 10 volt.

According to Ohm's Law -

$\large\sf{V = IR}$

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

$\large\sf{I = \frac{10}{9}\:A}$

We know that in series combination, the current flowing through each resistor is same and potential difference is different.

Therefore, for R1 -

$\large\sf{V1 = I \times R1}$

$\large\sf{V1 = \frac{10}{9} \times 2 }$

$\large\sf{\implies V1 = \frac{20}{9} \:volt}$

For R2 -

$\large\sf{ V2 = I \times R2}$

$\large\sf{ V2 = \frac{10}{9} \times 3}$

$\large\sf{\implies V2 = \frac{10}{3}\:volt}$

For R3 -

$\large\sf{ V3 = I \times R3}$

$\large\sf{V3 = \frac{10}{9} \times 4 }$

$\large\sf{ \implies V3 = \frac{40}{9} \:volt}$

Hence, the potential drop across 2 ohm, 3 ohm and 4 ohm resistor is $\sf{\frac{20}{9} \:volt ,\: \frac{10}{3}\:volt \:\:and\: \frac{40}{9}\:volt}$ respectively.