JEE MAIN - Physics (2021 - 1st September Evening Shift - No. 12)
In the given figure, each diode has a forward bias resistance of 30$$\Omega$$ and infinite resistance in reverse bias. The current I1 will be :
_1st_September_Evening_Shift_en_12_1.png)
3.75 A
2.35 A
2 A
2.73 A
Explanation
As per diagram,
Diode D1 & D2 are in forward bias i.e. R = 30$$\Omega$$ whereas diode D3 is in reverse bias i.e. R = infinite $$\Rightarrow$$ Equivalent circuit will be
Applying KVL starting from point A
_1st_September_Evening_Shift_en_12_2.png)
$$ - \left( {{{{I_1}} \over 2}} \right) \times 30 - \left( {{{{I_1}} \over 2}} \right) \times 130 - {I_1} \times 20 + 200 = 0$$
$$\Rightarrow$$ $$-$$100 I1 + 200 = 0
I1 = 2
Option (c)
Diode D1 & D2 are in forward bias i.e. R = 30$$\Omega$$ whereas diode D3 is in reverse bias i.e. R = infinite $$\Rightarrow$$ Equivalent circuit will be
Applying KVL starting from point A
$$ - \left( {{{{I_1}} \over 2}} \right) \times 30 - \left( {{{{I_1}} \over 2}} \right) \times 130 - {I_1} \times 20 + 200 = 0$$
$$\Rightarrow$$ $$-$$100 I1 + 200 = 0
I1 = 2
Option (c)
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