Q35 (2026)
Assuming in forward bias condition there is a voltage drop of 0.7 V across a silicon diode, the current through diode $D_{1}$ in the circuit is ____ mA. (Assume all diodes in the given circuit are identical)
{{IMG1}}
- $20.15$
- $11.7$
- $17.6$
- $18.8$
### Related Formula
$$I = \frac{V - V_d}{R}$$
### Core Logic
Check the polarity of the battery to determine which diodes are forward-biased. Diodes $D_1$ and $D_2$ are forward-biased, while $D_3$ is reverse-biased (acts as an open circuit). Because $D_1$ and $D_2$ are in parallel, the total voltage drop across the parallel combination is just $0.7 \mathrm{~V}$.
### Step 1: Loop Equation
Applying KVL to the main loop containing the forward-biased diodes:
$$12 - 0.3 \times 10^{3} I_{\text{total}} - 0.7 = 0$$
$$11.3 = 300 \cdot I_{\text{total}}$$
### Step 2: Total Current
$$I_{\text{total}} = \frac{11.3}{300} \mathrm{~A} = 37.66 \times 10^{-3} \mathrm{~A} = 37.66 \mathrm{~mA}$$
### Step 3: Current Division
Since $D_1$ and $D_2$ are identical and in parallel, the total current divides equally between them.
$$I_{D1} = \frac{I_{\text{total}}}{2} = \frac{37.66}{2} \mathrm{~mA} = 18.83 \mathrm{~mA}$$
Rounding to nearest option gives $18.8 \mathrm{~mA}$.
### Pattern Recognition
Identical diodes in parallel share the current equally. The voltage drop across the entire parallel diode bank is just the drop of one diode ($0.7 \mathrm{~V}$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Semiconductor Electronics: Materials, Devices and Simple Circuits
