JEE MAIN - Physics (2024 - 4th April Morning Shift - No. 17)
The value of net resistance of the network as shown in the given figure is :
_4th_April_Morning_Shift_en_17_1.png)
$$(30 / 11) \Omega$$
$$6 \Omega$$
$$(5 / 2) \Omega$$
$$(15 / 4) \Omega$$
Explanation
_4th_April_Morning_Shift_en_17_2.png)
Here $$D_1$$ is forward wise while $$D_2$$ is reversed wise
So net resistance between end, $$R=\frac{15 \times 10}{25}=6 \Omega$$
Comments (0)
