Question

The circuit shown contains two ideal diodes \(D_{1}\) and \(D_{2}\). If a cell of e.m.f. \(3\)V and negligible internal resistance is connected, then the current through \(70\Omega\) resistance in ampere is

• A. \(0.01\)
• B. \(0.02\)
• C. \(0.03\)
• D. \(0\)

Hint:

Check which diode is forward biased from the cell polarity, ignore the branch with the reverse biased diode, and apply Ohm's law.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
An ideal diode conducts with zero resistance when forward biased and acts as an open circuit when reverse biased.

Step 2: Check the Diodes
The positive terminal of the cell is on the left rail. Diode \(D_1\) has its cathode on that rail, so it is reverse biased and its branch carries no current.
Diode \(D_2\) has its anode on that rail, so it is forward biased.

Step 3: Find the Total Resistance
The active branch has \(30\,\Omega\) in series with the \(70\,\Omega\) resistor.
\(R=30+70=100\,\Omega\)

Step 4: Calculation
\(I=\dfrac{V}{R}=\dfrac{3}{100}=0.03\,\text{A}\), which is option (C).

Final Answer:
The current through the \(70\,\Omega\) resistor is 0.03 A. \[ \boxed{\text{(C) }0.03} \]