Question

Potential difference Vb-Va in the network shown is

Answer 1

Answer:

1 is answer for this question

Answer 2

Given that,

Resistance $R_{1}=2\ \Omega$

Resistance $R_{2}=1\ \Omega$

Resistance $R_{3}=2\ \Omega$

Resistance $R_{4}=1\ \Omega$

Resistance $R_{5}=3\ \Omega$

Potential = 12 V

potential = 3 V

According to figure,

R₁, R₂ and R₃ are connected in series

$R=R_{1}+R_{2}+R_{3}$

Put the value into the formula

$R=2+2+1$

$R=5\ \Omega$

R₄, and R₅ are connected in series

$R'=R_{4}+R_{5}$

Put the value into the formula

$R'=1+3$

$R'=4\ \Omega$

We need to calculate the equivalent emf

Using formula of equivalent emf

$\dfrac{E_{eq}}{R_{eq}}=\dfrac{E_{1}}{R_{1}}+\dfrac{E_{2}}{R_{2}}$

Put the value in to the formula

$\dfrac{E_{eq}}{R_{eq}}=\dfrac{12}{5}+\dfrac{3}{4}$

$\dfrac{E_{eq}}{R_{eq}}=\dfrac{48+15}{20}$

$\dfrac{E_{eq}}{R_{eq}}=\dfrac{63}{20}$....(I)

Resistance R and R' are connected in parallel

We need to calculate the equivalent resistance

Using formula of equivalent resistance

$\dfrac{1}{R_{eq}}=\dfrac{1}{R}+\dfrac{1}{R'}$

$\dfrac{1}{R_{eq}}=\dfrac{R'+R}{RR'}$

$R_{eq}=\dfrac{R\times R'}{R+R'}$

Put the value into the formula

$R_{eq}=\dfrac{4\times5}{4+5}$

$R_{eq}=\dfrac{20}{9}$

We need to calculate the potential difference

Using equation (I)

$\dfrac{E_{eq}}{R_{eq}}=\dfrac{63}{20}$

$E_{eq}=\dfrac{63}{20}\times\dfrac{20}{9}$

$E_{eq}=\dfrac{63}{9}$

$E_{eq}=7\ Volt$

Hence, The potential difference is 7 Volt.