A cell $\mathrm{E}_{1}$ of emf 6 V and internal resistance $2\Omega$ is connected with another cell $\mathrm{E}_{2}$ of emf 4 V and internal resistance $8\Omega$ (as shown in the figure). The potential difference across points $X$ and $Y$ is:

JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Current Electricity

A cell

E_{1}

of emf 6 V and internal resistance

2Ω

is connected with another cell

E_{2}

of emf 4 V and internal resistance

8Ω

(as shown in the figure). The potential difference across points

X

and

Y

is:

Answer: (C)

The two cells are connected in series with opposing polarities (based on typical circuit configuration).

Total emf in the loop:

E_{n e t} = E_{1} - E_{2} = 6 - 4 = 2

V (assuming

E_{1}

dominates)

Total resistance:

R_{t o t a l} = r_{1} + r_{2} = 2 + 8 = 10Ω

Current in circuit:

I = \frac{E _{n e t}}{R _{t o t a l}} = \frac{2}{10} = 0.2

A

Potential difference across

X

and

Y

(across cell

E_{1}

V_{X Y} = E_{1} - I r_{1} = 6 - 0.2 \times 2 = 6 - 0.4 = 5.6

V

The answer is (C) 5.6 V.

JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Single correct option