A cell E 1 of emf 6 V and internal resistance 2 Ω is connected with another cell E 2 of emf 4 V and internal resistance 8 Ω (as shown in the figure). The potential difference across points X and Y is

Question

A cell E 1 of emf 6 V and internal resistance 2 Ω is connected with another cell E 2 of emf 4 V and internal resistance 8 Ω (as shown in the figure). The potential difference across points X and Y is (A) 10.0 V (B) 3.6 V (C) 5.6 V (D) 2.0 V

Tardigrade Admin · Accepted Answer

I =(6-4/10)=(1/5) A V x +4+8 × (1/5)- V y =0 V x - V y =-5.6 ⇒ | Vx - Vy |=5.6 V

Q. 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

$10.0 V$

$3.6 V$

$5.6 V$

$2.0 V$

$I = \frac{6 - 4}{10} = \frac{1}{5} A$
$V_{x} + 4 + 8 \times \frac{1}{5} - V_{y} = 0$
$V_{x} - V_{y} = - 5.6$
$\Rightarrow ∣ V x - V y ∣ = 5.6 V$

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

10.0V

3.6V

5.6V

2.0V