Question

A p -n junction diode is connected to a battery of emf 5.7 V in series with a resistant 5kΩ such that it is forward biased. If the barrier potential of the diode is 0.7 V, neglecting the diode resistance, the current in the circuit is

• A. 1.14 mA
• B. 1mA
• C. 1A
• D. 1.14A

Answer 1

The Correct Option is B

In a forward biased p-n junction diode, the voltage across the diode is reduced by the barrier potential. The net voltage across the resistor in the circuit is the difference between the battery emf and the barrier potential of the diode.
Given:
Battery emf (\( V_{\text{battery}} \)) = 5.7 V
Barrier potential (\( V_{\text{barrier}} \)) = 0.7 V
Resistance (\( R \)) = 5 k\(\Omega\) = 5000 \(\Omega\)
The voltage across the resistor is: \[ V_{\text{resistor}} = V_{\text{battery}} - V_{\text{barrier}} = 5.7 \, \text{V} - 0.7 \, \text{V} = 5 \, \text{V} \] Using Ohm's law, \( V = IR \), the current in the circuit is: \[ I = \frac{V_{\text{resistor}}}{R} = \frac{5 \, \text{V}}{5000 \, \Omega} = 1 \, \text{mA} \] Thus, the current in the circuit is 1 mA.

Therefore, the correct answer is (B) 1 mA. 

Answer 2

In the given circuit, the battery provides a potential difference of 5.7 V, and the diode has a barrier potential of 0.7 V. Since the diode is forward biased, the voltage across the diode will be 0.7 V. The total potential available for the current to flow through the resistor is: \[ V_{\text{resistor}} = V_{\text{battery}} - V_{\text{diode}} = 5.7 \, \text{V} - 0.7 \, \text{V} = 5 \, \text{V}. \] Now, using Ohm's Law, the current \( I \) through the resistor (and hence through the diode) is: \[ I = \frac{V_{\text{resistor}}}{R} = \frac{5 \, \text{V}}{5000 \, \Omega} = 0.001 \, \text{A} = 1 \, \text{mA}. \] Thus, the current in the circuit is \( {1 \, \text{mA}} \).