Question

A cell E₁ of emf 6 V and internal resistance 2 Ω is connected with another cell E₂ of emf 4 V and internal resistance 8 Ω (as shown in the figure). The potential difference across points X and Y is ______.

Options

• 3.6 V

• 10.0 V

• 5.6 V

• 2.0 V

Answer 1

A cell E₁ of emf 6 V and internal resistance 2 Ω is connected with another cell E₂ of emf 4 V and internal resistance 8 Ω (as shown in the figure) . The potential difference across points X and Y is 5.6 V.

Explanation:

E₁ = 6 V

r₁ = 2 Ω 

E₂ = 4 V

r₂ = 8 Ω 

R_eq = r₁ +r₂ = 8 Ω + 2 Ω 

= 10 Ω 

E_eq = E₁ - E₂ 

E_eq = (6 - 4) V

E_eq =  2 V

I = `"E"_("eq")/"R"_"eq" = (2  "V")/(10Ω)`

I = 0.2 A

Hence, the potential difference across X and Y

V = E₂ + Ir₂    ...(∴ direction of flow of cWTent is anti clock)

⇒ V = 4 + 0.2 × 8

⇒ V = (4 + 1.6) V

⇒ 5.6 V