JEE MAIN - Physics (2021 - 26th February Morning Shift - No. 22)
The circuit contains two diodes each with a forward resistance of 50$$\Omega$$ and with infinite reverse resistance. If the battery voltage is 6V, the current through the 120$$\Omega$$ resistance is ____________ mA.
_26th_February_Morning_Shift_en_22_1.png)
Answer
20
Explanation
Given, forward resistance, R1 = 50 $$\Omega$$
Reverse resistance, R2 = infinity
Battery voltage = 6V
According to circuit diagram,
_26th_February_Morning_Shift_en_22_2.png)
In this case, diode D1 is forward biased, whereas diode D2 is reverse biased.
So, D2 will act as open circuit.
$$6 - 50I - 130I - 120I = 0$$
$$ \Rightarrow 6 = 300I$$
$$ \Rightarrow I = {6 \over {300}} = {1 \over {50}}$$
$$ = {2 \over {100}} = 0.02$$ A = 20 mA
Hence, current through 120 $$\Omega$$ = 20 mA
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