In the given figure, each diode has a forward bias resistance of 30 $$\Omega$$ and infinite resistance in reverse bias. The current $$I_1$$ will be:

Step-by-Step Circuit Solution

Based on the circuit diagram provided in the image, here is the clear, step-by-step breakdown of how to solve this question:

1. Identify the Diode Bias Conditions

The $$200\text{ V}$$ battery has its positive terminal on the left (the longer vertical line). The total current $$I_1$$ flows out from this positive terminal and travels upward into the three parallel branches:

• Top Branch: The diode points to the right ($$\rightarrow$$). Since current enters its anode (the triangle side), it is Forward Biased and conducts current.
• Middle Branch: This diode also points to the right ($$\rightarrow$$). Current enters its anode, so it is also Forward Biased and conducts current.
• Bottom Branch: This diode points to the left ($$\leftarrow$$). Current hits its cathode (the flat vertical bar side), making it Reverse Biased. It offers infinite resistance and acts as an open circuit (no current flows here).

2. Calculate the Resistance of Active Branches

Each forward-biased diode has a resistance of $$30\ \Omega$$.

• Top Branch Resistance:
  $$R_{\text{top}} = 30\ \Omega \text{ (diode)} + 130\ \Omega \text{ (resistor)} = 160\ \Omega$$
• Middle Branch Resistance:
  $$R_{\text{mid}} = 30\ \Omega \text{ (diode)} + 130\ \Omega \text{ (resistor)} = 160\ \Omega$$

3. Find the Total Equivalent Resistance ($$R_{\text{eq}}$$)

Since the top and middle branches are in parallel, we find their combined parallel resistance ($$R_p$$):

$$R_p = \frac{160 \times 160}{160 + 160} = 80\ \Omega$$

This parallel combination is connected in series with the $$20\ \Omega$$ resistor near the battery:

$$R_{\text{eq}} = R_p + 20\ \Omega = 80\ \Omega + 20\ \Omega = 100\ \Omega$$

4. Compute the Total Current ($$I_1$$)

Using Ohm's Law ($$I = \frac{V}{R}$$):

$$I_1 = \frac{200\text{ V}}{100\ \Omega} = 2.0\text{ A}$$

Correct Option: A (2.0 A)