Question 247: Current Electricity

Question

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

Accepted Answer

The circuit can be shown as below
The current through the circuit,
$$I = {{{\varepsilon _1} - {\varepsilon _2}} \over {{r_1} + {r_2}}} = {{6 - 4} \over {10}} = {1 \over 5}A$$
$$	herefore$$ Potential difference across points X and Y is
$${V_{XY}} = {E_2} + I{r_2} = 4 + {1 \over 5} 	imes 8 = 5.6V$$

Answer

10.0 V

Answer

2.0 V

Answer

5.6 V

Answer

3.6 V

Current Electricity

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

Choose the correct answer: