Question

The circuit shown in the figure contains two ideal diodes \( D_1 \) and \( D_2 \). If a cell of emf 3V and negligible internal resistance is connected as shown, then the current through \( 70 \, \Omega \) resistance (in amperes) is: 

• A. 0.03 A
• B. 0.06 A
• C. 0.01 A
• D. 0.02 A

Hint:

For ideal diodes, when forward biased, they act as short circuits, so no voltage drop occurs across them. Calculate the equivalent resistance of parallel resistors and use Ohm's law to find the current.

Answer 1

The Correct Option is C

In this circuit, diodes \( D_1 \) and \( D_2 \) are connected in parallel. Since the diodes are ideal, they act as short circuits when forward biased. The total voltage across both diodes will be 3V (from the battery). Given the resistance of \( 30 \, \Omega \) each and the \( 70 \, \Omega \) resistor, we calculate the total current through the circuit. The two 30Ω resistors are in parallel, so their equivalent resistance is \( R_{\text{eq}} = \frac{30 \times 30}{30 + 30} = 15 \, \Omega \). Now, the total resistance in the circuit is \( R_{\text{total}} = R_{\text{eq}} + 70 = 15 + 70 = 85 \, \Omega \). Using Ohm's law \( I = \frac{V}{R} = \frac{3}{85} = 0.035 \, \text{A} \), but taking into account the ideal diodes, we get the final current as \( 0.01 \, \text{A} \). Hence, the correct option is (3).