A diode has potential drop of 0.5 volts in forward bias. The maximumcurrent that can be flow through diode is 10mA.Then find the resistance connected in series with diode so that set up can be connected to a battery of 1.5 volts:

Options

A. `100ohm`
B. `50ohm`
C. `25ohm`
D. `10ohm`

Answer

To solve the problem step by step, we need to find the resistance connected in series with the diode when it is forward biased with a battery of 1.5 volts.

### Step 1: Identify the voltage drop across the diode
The diode has a potential drop of 0.5 volts when it is in forward bias.

### Step 2: Calculate the voltage across the resistor
The total voltage provided by the battery is 1.5 volts. Since the diode drops 0.5 volts, the voltage across the resistor (V_R) can be calculated as:
\[ V_R = V_{battery} - V_{diode} \]
Substituting the values:
\[ V_R = 1.5 \, \text{V} - 0.5 \, \text{V} = 1.0 \, \text{V} \]

### Step 3: Identify the maximum current through the diode
The maximum current that can flow through the diode is given as 10 mA (or 0.01 A).

### Step 4: Use Ohm's Law to find the resistance
Ohm's Law states that:
\[ V = I \cdot R \]
Where:
- \( V \) is the voltage across the resistor,
- \( I \) is the current through the resistor,
- \( R \) is the resistance.

We can rearrange this formula to solve for resistance:
\[ R = \frac{V}{I} \]
Substituting the values we found:
\[ R = \frac{1.0 \, \text{V}}{0.01 \, \text{A}} = 100 \, \Omega \]

### Conclusion
The resistance that should be connected in series with the diode is **100 ohms**.

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