Question

The value of net resistance of the network as shown in the given figure is :

• A. \( \frac{5}{2} \, \Omega \)
• B. \( \frac{15}{4} \, \Omega \)
• C. \( 6 \, \Omega \)
• D. \( \frac{30}{11} \, \Omega \)

Answer 1

The Correct Option is C

The equivalent resistance is calculated as follows:

\[ R_{\text{eq}} = \frac{15 \times 10}{15 + 10} = \frac{150}{25} = 6 \, \Omega \]

 Diode 2 is in reverse bias, so no current will flow through the branch containing it. It can be treated as a broken wire.

Diode 1 is in forward bias and will behave as a conducting wire. The new circuit becomes a simpler parallel and series resistor network.

Answer 2

Step 1: Understanding the given circuit.
The circuit contains three resistors — 15 Ω, 10 Ω, and 5 Ω — connected between two terminals at potentials of −6 V and −8 V.
The 10 Ω and 5 Ω resistors each have diodes connected in opposite directions.

Step 2: Direction of current flow.
Since −6 V is at a higher potential than −8 V, current flows from left to right.

Step 3: Behavior of diodes.
- In the 10 Ω branch, the diode is forward-biased (conducting) because it allows current from left to right.
- In the 5 Ω branch, the diode is reverse-biased (non-conducting) and blocks current.
- The 15 Ω branch has no diode and therefore conducts normally.

Step 4: Simplify the circuit.
Only the 15 Ω and 10 Ω resistors conduct current, and they are connected in parallel.
The equivalent resistance is:
\[ \frac{1}{R_{\text{eq}}} = \frac{1}{15} + \frac{1}{10} \] \[ \frac{1}{R_{\text{eq}}} = \frac{2 + 3}{30} = \frac{5}{30} = \frac{1}{6} \] \[ R_{\text{eq}} = 6 \, \Omega \]
Step 5: Final Answer.
\[ \boxed{R_{\text{eq}} = 6 \, \Omega} \]