Question

A battery of 9V is connected in series with resistors of 2 ohms, 3 ohms and 5 ohms. Calculate the potential difference across 3 ohms

Answer 1

as current is same in all resistor therefore

V1:V2:V3=R1:R2:R3

therefore V1:V2:V3=2:3:5

therefore V2= (ratio of V2/sum of ratio)× V total therefore

V2=(3/10)×9= 2.7v

...

hope it helpful

Answer 2

★ Given :-

• R 1 = 2 Ω
• R 2 = 3 Ω

• R 3 = 5 Ω

• Potential difference  = 9V

★ To find :-

• Potential difference across 3 Ω resistor

★ Solution :-

First we need to find the equivalent resistance of the circuit

Equivalent resistance for three  resistors connected in Series is given by the formula ,

$\bigstar\boxed{\mathtt{ R_s = R_1 + R_2 ....+ R_n }}$

hence ,

$\implies\mathtt{ R_s = R_1 + R_2+ R_3 }$

$\implies\mathtt{ R_s = 2 +3+5 }$

$\implies\mathtt{ R_s = 10 \Omega}$

The equivalent resistance of the circuit is 10Ω

Now let's find the net current flowing through the circuit in order to do that simply apply ohm's law

According to ohm's law ,

$\bigstar\boxed{\mathtt{ V= IR }}$

hence ,

$\implies\mathtt{ I = \dfrac{V}{R} }$

$\implies\mathtt{ I = \dfrac{9}{10} }$

$\implies\mathtt{ I =0.9A }$

The current flowing through the circuit is 0.9 A

We know that when  resistors are connected in series with each other the potential difference across the resistors is different but the current flowing through  these resistors is same

Potential difference across 3Ω resistor

$\implies\mathtt{ V = I R }$

$\implies\mathtt{ V = 0.9 \times 3 }$

$\implies\mathtt{ V = 2.7 V }$

The potential difference across 3Ω resistor is 2.7  volt