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Question: In the given figure a diode D is connected to an external resistance \(R = 100\Omega \)and an emf of...

In the given figure a diode D is connected to an external resistance R=100ΩR = 100\Omega and an emf of 3.5V3.5V. If the barrier potential developed across the diode is 0.5V0.5V, the current in the circuit will be

A. 35mA35mA
B. 30mA30mA
C. 40mA40mA
D. 20mA20mA

Explanation

Solution

potential across diode is given it means diode has some internal resistance use kirchhoff's voltage law or KVL to find the current in the circuit( Remember in forward biased diode do not oppose the current so it will acts as a plane wire but due to some internal resistance of the diode we may see some potential difference across it).

Complete step by step answer:
Let current in the circuit be AmpereAmpere and
Internal resistance of the diode is RΩR\Omega
Now total resistance of the circuit will be sum of external resistance and internal resistance of the diode
Therefore,Rtotal=(R+100){R_{total}} = (R + 100)
Now we will use ohm's law across whole circuit and across diode
For diode
We have Vdiode=0.5=IR{V_{diode}} = 0.5 = IR
For whole circuit
We have Vtotal=3.5=I(R+100){V_{total}} = 3.5 = I(R + 100)
3.5=IR+I×100\Rightarrow 3.5 = IR + I \times 100
Putting value of IR from the potential equation of diode
We have ,
3.5=0.5+I×100 I=3.50.5100=2100=20×103  3.5 = 0.5 + I \times 100 \\\ \Rightarrow I = \dfrac{{3.5 - 0.5}}{{100}} = \dfrac{2}{{100}} = 20 \times {10^{ - 3}} \\\
Or I=20mAI = 20mA

So, the correct answer is “Option D”.

Note:
Sometimes students are scared just looking at the diode because they understand the diode properly in most cases but even if you do not understand it properly if you look at the type of bias it becomes like a normal circuit if its its forward biased treat diode as a plane wire with some internal resistance and if its reverse biased treat it as a open wire.