JEE MAIN - Physics (2021 - 24th February Morning Shift - No. 7)
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 :
_24th_February_Morning_Shift_en_7_1.png)
10.0 V
2.0 V
5.6 V
3.6 V
Explanation
The circuit can be shown as below
_24th_February_Morning_Shift_en_7_2.png)
The current through the circuit,
$$I = {{{\varepsilon _1} - {\varepsilon _2}} \over {{r_1} + {r_2}}} = {{6 - 4} \over {10}} = {1 \over 5}A$$
$$\therefore$$ Potential difference across points X and Y is
$${V_{XY}} = {E_2} + I{r_2} = 4 + {1 \over 5} \times 8 = 5.6V$$
The current through the circuit,
$$I = {{{\varepsilon _1} - {\varepsilon _2}} \over {{r_1} + {r_2}}} = {{6 - 4} \over {10}} = {1 \over 5}A$$
$$\therefore$$ Potential difference across points X and Y is
$${V_{XY}} = {E_2} + I{r_2} = 4 + {1 \over 5} \times 8 = 5.6V$$
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