Question

In the circuit shown, if each diode has a forward biased resistance of 75 ohm and infinite resistance bias, what will be the current in i1

Answer 1

Answer:

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Explanation:

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Answer 2

Given: The circuit is given and the forward-biased resistance is $25$ Ω

To find: The current $i_{1}$

Solution:

The branch in the diode is reverse biased in current $i_{3}$ so it is zero

Resistance in the branch AB and EF are each= $75$ Ω$+$$25$ Ω=$100$ Ω

The branches AB and EF are identical parallel branches,

so their effective resistance is

$=\dfrac{100}{2}$

$=50$ Ω

Then, the Net resistance of the circuit is  $=(50+75)$

$=125$ Ω

Current

$I_{1}=\dfrac{15}{125}$

$=0.12$ $A$

The resistances of AB and EF are equal,

$I_{1}=I_{2}+I_{3}+I_{4}$

as $I_{3}=0$

$I_{2}=I_{4}=\dfrac{0.12}{2}$

$I_{2}=I_{4}=\dfrac{0.12}{2}\\=0.6A$

Therefore, the currents are $I_{1} =0.12 A, I_{2}= 0.6, I_{3}=0$ and $I_{4}=0.6 A$