In the given figure, each diode has a forward bias resistance of 30 $Ω$ and infinite resistance in reverse bias. The current I₁ will be:

Option 1 -

3.75 A

Option 2 -

2.73 A

Option 3 -

2.35 A

Option 4 -

2 A

5 Views | Posted 2 months ago

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1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 4

Detailed Solution:

In give diagram Diode 1 and 2 are in forward bias with R = $30 Ω$ and Diode 3 is reverse bias with R = infinite l₁ current is flowing through 20 $Ω$

So $\frac{l_{1}}{2} a n d \frac{l_{1}}{2}$ current will flow through Diode 1 and 2. As resistance is same applying KCl in ABCD Loop $- \frac{l_{1}}{2} \times 130 - \frac{l_{1}}{2} \times 130 - l_{1} \times 20 + 200 = 0$

-100l₁ + 200 = 0

l₁ = 2

In give diagram Diode 1 and 2 are in forward bias with R = <math> <mrow> <mn>3</mn> <mn>0</mn> <mi>Ω</mi> </mrow> </math> <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1025" DrawAspect="Content" ObjectID="_1815996175"> </o:OLEObject></xml><! [endif]->and Diode 3 is reverse bias with R = infinite l1 current is flowing through 20 <math> <mrow> <mi>Ω</mi> </mrow> </math>  <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1026" DrawAspect="Content" ObjectID="_1815996176"> </o:OLEObject></xml><! [endif]->So <math> <mrow> <mfrac> <mrow> <msub> <mrow> <mi>l</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mi>a</mi> <mi>n</mi> <mi>d</mi> <mtext> </mtext> <mtext> </mtext> <mfrac> <mrow> <msub> <mrow> <mi>l</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> </math> <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1027" DrawAspect="Content" ObjectID="_1815996177"> </o:OLEObject></xml><! [endif]->current will flow through Diode 1 and 2. As resistance is same applying KCl in ABCD Loop <math> <mrow> <mo>−</mo> <mfrac> <mrow> <msub> <mrow> <mi>l</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>×</mo> <mn>1</mn> <mn>3</mn> <mn>0</mn> <mo>−</mo> <mfrac> <mrow> <msub> <mrow> <mi>l</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>×</mo> <mn>1</mn> <mn>3</mn> <mn>0</mn> <mo>−</mo> <msub> <mrow> <mi>l</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>×</mo> <mn>2</mn> <mn>0</mn> <mo>+</mo> <mn>2</mn> <mn>0</mn> <mn>0</mn> <mo>=</mo> <mn>0</mn> </mrow> </math> -100l1 + 200 = 0l1 = 2

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In the given figure, each diode has a forward bias resistance of 30 $Ω$ and infinite resistance in reverse bias. The current I₁ will be:

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In the given figure, each diode has a forward bias resistance of 30 Ω and infinite resistance in reverse bias. The current I1 will be:

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In the given figure, each diode has a forward bias resistance of 30 $Ω$ and infinite resistance in reverse bias. The current I₁ will be: