Question

Identify which of the two diodes will conduct and why ?

Answer 1

Concept: An ideal diode conducts only when it is forward biased and behaves like a closed switch. When it is reverse biased, it behaves like an open switch and no current flows through it. For a diode:
• Forward bias : Anode at a higher potential than the cathode.
• Reverse bias : Cathode at a higher potential than the anode.

Step 1: Determine the polarity during the positive half-cycle.
During the positive half-cycle of the input voltage \(V_i\), \[ V_A>V_B, \] that is, point \(A\) is at a higher potential than point \(B\).

Step 2: Examine the biasing of the two diodes.
From the figure:
• For diode \(D_1\), the cathode is connected towards point \(A\). Since \(A\) is at a higher potential, \(D_1\) becomes reverse biased.
• For diode \(D_2\), the anode is connected towards point \(A\) and its cathode is towards point \(R\). Therefore, \(D_2\) becomes forward biased. Hence, \[ \boxed{\text{Diode } D_2 \text{ conducts and diode } D_1 \text{ remains OFF.}} \] The reason is that during the positive half-cycle, \(D_2\) is forward biased whereas \(D_1\) is reverse biased.

Answer 2

From part (a), during the positive half-cycle of the input voltage, diode \(D_2\) is forward biased and conducts, whereas diode \(D_1\) is reverse biased and does not conduct. Therefore,
• \(D_2\) is replaced by a conducting wire (short circuit).
• \(D_1\) is replaced by an open circuit. Hence, the equivalent circuit becomes \[ \begin{array}{c} A[0.2cm] | | R |\qquad\backslash 1\,k\Omega \qquad 3\,k\Omega |\qquad\qquad\backslash P \qquad\qquad B \backslash 2\,k\Omega \backslash B \end{array} \] Thus, the \(1\,k\Omega\) and \(2\,k\Omega\) resistors are connected in series between \(P\) and \(B\), and this series combination is in parallel with the \(3\,k\Omega\) resistor.

Answer 3

Concept: The applied voltage is \[ V_i=12\sin(100\pi t)\ \text{V}. \] The peak value of the input voltage is \[ V_{\text{peak}}=12\ \text{V}. \] From part (b), the equivalent circuit consists of:
• a \(3\,k\Omega\) resistor directly across the source,
• a series combination of \(1\,k\Omega\) and \(2\,k\Omega\) resistors also connected across the source. Hence, both branches have the same potential difference of \(12\) V.

Step 1: Calculate the current through the branch containing \(1\,k\Omega\) and \(2\,k\Omega\).
The equivalent resistance of this branch is \[ R_s=1\,k\Omega+2\,k\Omega=3\,k\Omega. \] Therefore, \[ I=\frac{12}{3\times10^3} =4\times10^{-3}\ \text{A} =4\ \text{mA}. \]

Step 2: Calculate the voltage drop across the \(1\,k\Omega\) resistor.
::contentReference[oaicite:0]{index=0} \[ V_{1k}=IR \] \[ V_{1k} = (4\times10^{-3})(1000) = 4\ \text{V}. \] Hence, \[ \boxed{V_{1k}=4\ \text{V}} \]

Step 3: Calculate the voltage drop across the \(2\,k\Omega\) resistor.
\[ V_{2k} = (4\times10^{-3})(2000) = 8\ \text{V}. \] Therefore, \[ \boxed{V_{2k}=8\ \text{V}} \]

Step 4: Calculate the voltage drop across the \(3\,k\Omega\) resistor.
Since the \(3\,k\Omega\) resistor is directly connected across the source, \[ V_{3k}=12\ \text{V}. \] Thus, \[ \boxed{V_{3k}=12\ \text{V}} \] Hence, the output voltage drops across the three resistors are \[ \boxed{ V_{1k}=4\ \text{V},\qquad V_{2k}=8\ \text{V},\qquad V_{3k}=12\ \text{V} } \]