If each diode in figure has a forward ha...

If each diode in figure has a forward has a forward bias resistance of 25 Omega and infinite resistance in reverse bias, what will be the values of the current ` I_1 , I_2 , I_3 and I_4? `

`I_2=0.40 A, I_4=0.025 A`

`I_2=0.25 A , I_4=0.20 A`

`I_1=0.05 A , I_3=0.02 A`

`I_2=I_4=0.025 A`

Text Solution

Verified by Experts

The correct Answer is:

Let R be the effective resistance of the circuit, then
`R=(R_(AB)||R_(EF))+25`
`R_(AB)=125+25=150 Omega, R_(EF)=125+25=150 Omega`
`therefore R=25+150/2=100 Omega`
Since diode in the branch CD is reverse biased, `I_3=0`
Current, `I_1=5/100=0.05 A`
According to Kirchhoff's current rule,
`I_1=I_2+I_3+I_4 " " or " " I_2+I_4=I_1=0.05 A`
`because R_(AB)=R_(EF),so, I_4=I_2`
`2I_4=2I_4=0.05 , I_4=I_2=0.05/2=0.025 A`

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