Question

In the given circuit, the voltage across load resistance (\( R_L \)) is:

• A. 8.75 V
• B. 9.00 V
• C. 8.50 V
• D. 14.00 V

Answer 1

The Correct Option is A

Given Data:

R_D1 = 1.5 kΩ,   R_L = 2.5 kΩ,   V = 15 V,   V_D1 = 0.3 V,   V_D2 = 0.7 V.

The voltage across the resistive portion of the circuit can be found by subtracting the voltage drops across the two diodes from the applied voltage:

V_resistive = V - (V_D1 + V_D2).

Substituting the values:

V_resistive = 15 V - (0.3 V + 0.7 V) = 14 V.

The total resistance in the resistive portion of the circuit is the sum of R_D1 and R_L:

R_total = R_D1 + R_L = 1.5 kΩ + 2.5 kΩ = 4.0 kΩ.

Using Ohm’s Law, the total current i in the circuit is given by:

i = \(\frac{V_{resistive}}{R_{total}}\).

Substituting the values:

i = \(\frac{14 \text{ V}}{4.0 \text{ k}\Omega}\) = \(\frac{14}{4000}\) A = 0.0035 A = 3.5 mA.

The voltage across the load resistance R_L, denoted as V_L, is given by Ohm’s Law:

V_L = i × R_L.

Substituting the values:

V_L = 3.5 mA × 2.5 kΩ = 0.0035 A × 2500 Ω = 8.75 V.

Final Result: The voltage across the load resistance R_L is:

V_L = 8.75 V

Answer 2

The problem asks to determine the voltage across the load resistance \(R_L\) in a series circuit containing a voltage source, two different types of diodes, and a series resistor.

Concept Used:

To solve this circuit, we will use the following concepts:

1. Diode Forward Biasing: When the positive terminal of a voltage source is connected to the anode and the negative terminal to the cathode, a diode is forward-biased and allows current to flow.
2. Diode Barrier Potential (Forward Voltage Drop): In a practical model, a forward-biased diode introduces a small, constant voltage drop across it. This voltage drop is known as the barrier potential.
   • For a Germanium (Ge) diode, the barrier potential is approximately \(V_{Ge} \approx 0.3 \, \text{V}\).
   • For a Silicon (Si) diode, the barrier potential is approximately \(V_{Si} \approx 0.7 \, \text{V}\).
3. Kirchhoff's Voltage Law (KVL): The algebraic sum of all the voltages around any closed loop in a circuit is equal to zero. This means the total voltage supplied by the source is equal to the sum of the voltage drops across all components in the series loop.
4. Ohm's Law: The voltage drop (\(V\)) across a resistor is directly proportional to the current (\(I\)) flowing through it, given by \(V = IR\).

Step-by-Step Solution:

Step 1: Analyze the circuit and check the biasing of the diodes.

The circuit consists of a 15V source connected in series with a Ge diode (D₁), a Si diode (D₂), a 1.5 kΩ resistor, and a 2.5 kΩ load resistor (R_L). The positive terminal of the 15V source is connected to the anode of D₁, and the subsequent components are in series. This configuration means both diodes are forward-biased and will conduct current.

Step 2: Apply Kirchhoff's Voltage Law (KVL) to the circuit loop.

The total voltage from the source (\(V_s\)) will be dropped across the two diodes, the series resistor (\(R_S\)), and the load resistor (\(R_L\)). Let \(I\) be the current flowing through the series circuit.

\[ V_s = V_{Ge} + V_{Si} + V_{R_S} + V_{R_L} \] \[ V_s = V_{Ge} + V_{Si} + I \cdot R_S + I \cdot R_L \] \[ V_s = V_{Ge} + V_{Si} + I (R_S + R_L) \]

Step 3: Calculate the net voltage available for the resistive part of the circuit.

The total voltage drop across the diodes is the sum of their barrier potentials.

\[ V_{\text{diodes}} = V_{Ge} + V_{Si} = 0.3 \, \text{V} + 0.7 \, \text{V} = 1.0 \, \text{V} \]

The remaining voltage from the source is dropped across the resistors. Let's call this \(V_{\text{resistors}}\).

\[ V_{\text{resistors}} = V_s - V_{\text{diodes}} = 15 \, \text{V} - 1.0 \, \text{V} = 14 \, \text{V} \]

Step 4: Calculate the total resistance of the circuit and the circuit current (\(I\)).

The total resistance in the circuit is the sum of the series resistor and the load resistor.

\[ R_{\text{total}} = R_S + R_L = 1.5 \, \text{k}\Omega + 2.5 \, \text{k}\Omega = 4.0 \, \text{k}\Omega = 4000 \, \Omega \]

The current \(I\) can now be calculated using Ohm's Law with the net voltage across the resistors.

\[ I = \frac{V_{\text{resistors}}}{R_{\text{total}}} = \frac{14 \, \text{V}}{4.0 \, \text{k}\Omega} = \frac{14}{4000} \, \text{A} = 0.0035 \, \text{A} = 3.5 \, \text{mA} \]

Final Computation & Result:

Step 5: Calculate the voltage across the load resistance (\(R_L\)).

Using Ohm's Law, the voltage across the load resistance \(R_L\) is:

\[ V_{R_L} = I \times R_L \]

Substitute the values of the current \(I\) and resistance \(R_L\):

\[ V_{R_L} = (3.5 \times 10^{-3} \, \text{A}) \times (2.5 \times 10^3 \, \Omega) \] \[ V_{R_L} = 3.5 \times 2.5 \, \text{V} \] \[ V_{R_L} = 8.75 \, \text{V} \]

The voltage across the load resistance \(R_L\) is 8.75 V.