Question

Find $V_{AB}$ in the circuit given below.

Hint:

The diode is forward biased, so treat it as a short. Then two 10 ohm resistors in parallel (5 ohm) are in series with 5 ohm, and V_AB is half of 15 V.

Answer 1

Step 1: Understanding the Concept:
The circuit has a 15 V battery, a 5 ohm resistor, and two parallel branches between A and B. One branch is a 10 ohm resistor. The other branch is an ideal diode in series with another 10 ohm resistor. We first decide if the diode conducts.

Step 2: State of the diode:
The longer plate of the battery (the positive terminal) is connected through the 5 ohm resistor to point A. So A is at a higher potential than B.
The diode has its arrow pointing from A towards D, i.e. from the high potential side to the low side. It is forward biased and acts as a short circuit (ideal diode).

Step 3: Equivalent resistance:
The two 10 ohm resistors are now in parallel between A and B:
\[ R_{AB} = \frac{10 \times 10}{10 + 10} = 5\ \Omega \]
This is in series with the 5 ohm resistor, so the total resistance is \(5 + 5 = 10\ \Omega\).

Step 4: Current and voltage:
\[ I = \frac{15}{10} = 1.5\ \text{A} \]
\[ V_{AB} = I \times R_{AB} = 1.5 \times 5 = 7.5\ \text{V} \]

Final Answer:
The diode conducts and the voltage between A and B is 7.5 V. \[ \boxed{V_{AB} = 7.5\ \text{V}} \]