In the circuit shown in figure all the three diodes D1, D2, D3 have forward resistance \(50 \Omega\) each and infinite backward resistance. If the battery voltage is 5V, the current through \(100 \Omega\) resistance is
Show Hint
Find which diodes conduct, then combine the resistances.
Step 1: Understanding the Concept
A forward-biased diode here has resistance \(50\ \Omega\). A reverse-biased diode has infinite resistance and blocks current.
Step 2: Key Formula or Approach
From the figure, the battery's positive terminal feeds D1 and D2 in the forward direction, while D3 points the opposite way, so it is reverse biased and carries no current.
Step 3: Detailed Explanation
Branch 1: \(D_1\) (50) + 150 = \(200\ \Omega\).
Branch 2: \(D_2\) (50) + 50 = \(100\ \Omega\).
Parallel: \(\dfrac{200\times100}{300}=\dfrac{200}{3}\ \Omega\).
Add the 100 ohm series resistor: \(R_{tot}=100+\dfrac{200}{3}=\dfrac{500}{3}\ \Omega\).
\[ I=\frac{5}{500/3}=0.03\ \text{A}=30\ \text{mA} \]
The same current flows through the 100 ohm resistor.
Final Answer:
The current through the 100 ohm resistor is 30 mA, option (B).
\[ \boxed{30\ \text{mA (B)}} \]
