The circuit shown in the figure, the forward voltage of the diode is 0.7 V and its dynamic resistance is 2 \(\Omega\). The current through the 20 \(\Omega\) resistor is
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When analyzing diode circuits with a more complex model (including dynamic resistance), replace each forward-biased diode with its equivalent circuit (a voltage source in series with a resistor) and then solve the resulting linear circuit using standard methods like KVL.
Step 1: Analyze the DC circuit. The 5V source powers a diode bridge. Current flows from the positive terminal, splitting into paths. A 20 \(\Omega\) resistor lies in one path. The current's route is: positive source terminal, diode D1, 20 \(\Omega\) resistor, diode D3, negative source terminal.
Step 2: Apply KVL. The loop includes the 5V source, diode D1, 20 \(\Omega\) resistor, and diode D3.
\[ V_{source} - V_{D1} - I \cdot R_{20\Omega} - V_{D3} = 0 \]
Step 3: Model the forward-biased diodes. Each diode is modeled as a 0.7V voltage source in series with a 2 \(\Omega\) dynamic resistance. Thus, \(V_{D1} = 0.7V + I \cdot (2\Omega)\) and \(V_{D3} = 0.7V + I \cdot (2\Omega)\).
Step 4: Calculate the current I. Substitute the diode models into the KVL equation:
\[ 5 - (0.7 + 2I) - 20I - (0.7 + 2I) = 0 \]
\[ 5 - 1.4 - 2I - 20I - 2I = 0 \]
\[ 3.6 - 24I = 0 \]
\[ 24I = 3.6 \]
\[ I = \frac{3.6}{24} = 0.15 \text{ A} \]
Therefore, \(I = 0.15 \text{ A} = 150 \text{ mA}\).
