For the circuit shown, with \(\mathrm{R}_{1}=1.0 \Omega, \mathrm{R}_{2}=2.0 \Omega, \mathrm{E}_{1}=2 \mathrm{~V}\) and \(\mathrm{E}_{2}=\mathrm{E}_{3}=4 \mathrm{~V}\), the potential difference between the points ' \(a\) ' and ' \(b\) ' is approximately (in V ):

JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Current Electricity

For the circuit shown, with

R_{1} = 1.0Ω, R_{2} = 2.0Ω, E_{1} = 2 V

and

E_{2} = E_{3} = 4 V

, the potential difference between the points '

a

' and '

b

' is approximately (in V ):

Answer: (A)

Treat the three branches between

a

and

b

as cells in parallel and use the equivalent-EMF idea, in which each branch contributes its EMF weighted by its own conductance:

E_{e q} = \frac{\frac{E _{1}}{2 R _{1}} + \frac{E _{2}}{R _{2}} + \frac{E _{3}}{2 R _{1}}}{\frac{1}{2 R _{1}} + \frac{1}{R _{2}} + \frac{1}{2 R _{1}}}

. Putting

R_{1} = 1.0 Ω

R_{2} = 2.0 Ω

E_{1} = 2

V and

E_{2} = E_{3} = 4

V, the numerator becomes

\frac{2}{2} + \frac{4}{2} + \frac{4}{2} = 5

and the denominator

\frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{3}{2}

. The potential difference between

a

and

b

is therefore

\frac{5}{3/2} = \frac{10}{3} \approx 3.3

JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Single correct option