Question

The following circuit contains diodes with forward bias having resistance \(25Ω\), and reverse bias having infinite resistance. The ratio of \(\frac{I_{1}}{I_{2}}\) is equal to

 

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The Correct Option is B

To solve the given problem, we need to analyze the diode circuit and calculate the ratio of \( \frac{I_{1}}{I_{2}} \).

Let's break down the steps:

1. The circuit consists of a voltage source of 5V, a resistor of \(25 \, \Omega\) in series with diodes and resistors of \(125 \, \Omega\) each.
2. Diodes conduct current only when they are forward biased. For simplicity, we consider a forward-biased diode as a resistor with \(25 \, \Omega\) resistance, and an open circuit (infinite resistance) when reverse-biased.
3. Since \(I_1\) passes through the \(25 \, \Omega\) resistor, \(I_2\) must pass through a combination of a diode and a resistor. We assume that if the diode is forward-biased, it will indeed conduct.
4. Considering ideal behavior of diodes, the \(\text{current } I_1\) will also pass through the \(25 \, \Omega\) resistor. If the diode is reverse-biased, the circuit path providing \(I_2\) will not conduct, thus \(I_2 = 0\).
5. However, assuming diodes can conduct since voltage across \(I_2\) path is zero for reverse-bias, the effective resistance over the path including a forward-biased diode is \(125 \, \Omega + 25 \, \Omega = 150 \, \Omega\).
6. Using Ohm’s law, we calculate currents assuming \(I_2\) has conducted path:

Ohm's Law: \( V = IR \)

The total voltage is distributed across the conducting paths. Therefore, using the concept of current division:

• \(\frac{I_{1}}{I_{2}} = \frac{R_2}{R_1} = \frac{150}{75} = 2\)

Hence, the ratio \( \frac{I_1}{I_2} \) is 2.

Therefore, the correct answer is option 2.