The circuit shown in the figure contains two diodes each with a forward resistance of `50Omega` and with infinite backward resistance. If the batter of 6 V is connected in the circuit, then the current through the `100 Omega` resistance is

Options

A. Zero
B. `0.02 A`
C. `0.03 A`
D. `0.036 A`

Answer

To solve the problem, we need to analyze the circuit containing two diodes, a 100-ohm resistor, and a 6V battery. Here’s a step-by-step solution:

### Step 1: Understand the Circuit Configuration
In the given circuit, we have:
- A 6V battery connected in series with a 100-ohm resistor.
- Two diodes (D1 and D2) connected in opposite directions.
- D1 has a forward resistance of 50 ohms, and D2 also has a forward resistance of 50 ohms.

### Step 2: Determine the Forward Bias Condition
Since the diodes are connected in opposite directions, only one of them will be forward-biased when the battery is connected. Given that the battery is 6V, we can assume that D1 is forward-biased and D2 is reverse-biased.

### Step 3: Calculate Total Resistance in the Circuit
The total resistance in the circuit can be calculated by adding the resistances in series:
- The forward resistance of D1 = 50 ohms
- The 100-ohm resistor = 100 ohms
- The forward resistance of D2 is not considered because it is reverse-biased (infinite resistance).

Thus, the total resistance (R_total) is:
\[ R_{\text{total}} = R_{100} + R_{D1} = 100 \, \Omega + 50 \, \Omega = 150 \, \Omega \]

### Step 4: Apply Ohm's Law to Find Current
Using Ohm's Law, which states \( I = \frac{V}{R} \), we can find the current (I) flowing through the circuit:
- Voltage (V) = 6V
- Total Resistance (R_total) = 150 ohms

Now, substituting the values:
\[ I = \frac{6V}{150 \, \Omega} = 0.04 \, A \]

### Step 5: Conclusion
The current through the 100-ohm resistor is:
\[ I = 0.04 \, A \text{ or } 40 \, mA \]