JEE MAIN - Physics (2024 - 8th April Morning Shift - No. 19)
In the given circuit, the terminal potential difference of the cell is :
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3 V
4 V
2 V
1.5 V
Објашњење
_8th_April_Morning_Shift_en_19_2.png)
$$V_T=\left(\frac{3}{1+2}\right) \times 2=2 \mathrm{~V}$$
In the given circuit, the terminal potential difference of the cell is :
$$V_T=\left(\frac{3}{1+2}\right) \times 2=2 \mathrm{~V}$$