Question

In the given circuit, the potential difference between A and B is:
 

\[A)0\]
\[B)5volt\]
\[C)10volt\]
\[D)15volt\]

Answer 1

Hint: When a diode is forward biased in a circuit, it acts as a conductor, hence the equivalent resistance across AB can be found. Once the equivalent resistance across the terminals A and B is determined, we can find the potential difference across AB.
Formula used:
\[\dfrac{1}{R}=\dfrac{1}{{{R}_{1}}}+\dfrac{1}{{{R}_{2}}}\]
\[V=IR\]

Complete answer:
Given, three resistors with \[10\Omega \] each are connected to the circuit. It has a voltage input of \[{{V}_{in}}=30V\]
Here, diode allows current, since it is forward biased. Therefore, the last two resistors have a parallel connection. Then,
Equivalent resistance across AB, \[\dfrac{1}{{{R}_{AB}}}=\dfrac{1}{10}+\dfrac{1}{10}=\dfrac{10+10}{10\times 10}=\dfrac{1}{5}\Omega \]
Then,
\[{{R}_{AB}}=5\Omega \]
Since the resistors are connected in series, the voltage gets divided across the resistors.
Voltage across \[10\Omega \] resistor, \[V=IR=10I\] --------1
Voltage across, \[5\Omega \]resistor,\[{{V}_{AB}}=I{{R}_{AB}}=5I\] -------2
Where, \[I\] is the current passing through the circuit.
Divide equation1 by 2. Then
\[\dfrac{V}{{{V}_{AB}}}=\dfrac{10I}{5I}=\dfrac{2}{1}\]
Hence, the potential gets divided in the ratio 2:1 across the first resistance \[10\Omega \] and the equivalent resistance across AB which is \[5\Omega \].
Hence, the potential difference across AB \[=\dfrac{1}{3}{{V}_{in}}=\dfrac{1}{3}\times 30=10V\]

So, the correct answer is “Option C”.

Additional Information:
Diodes are used in circuits to allow the current flow in one direction, while blocking the current flow in the opposite direction through a circuit. It converts the power from AC to DC. Also, it helps to regulate the voltage in a circuit.

Note:
Diodes can be either forward biased or reverse biased. If it is forward biased, i.e., if its p region is connected to the positive of the battery and n region is connected to the negative of the battery, it is called a forward biased diode. In this case it acts as a conductor. But when it is reverse biased, i.e., p region is connected to negative of the battery and n region to the positive of the battery. Then, it acts as a resistor, which limits the current flow through the circuit.