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 circuit paths for the DC source.
The 5V source is connected to a diode bridge. The current will flow from the positive terminal and split. The 20 \(\Omega\) resistor is in one of the paths through the bridge. The path for current through the resistor is from the positive terminal of the source, through diode D1, through the 20 \(\Omega\) resistor, through diode D3, to the negative terminal of the source.
Step 2: Apply Kirchhoff's Voltage Law (KVL) to the relevant loop.
The loop consists of the 5V source, diode D1, the 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.
The model for a forward-biased diode is a voltage source of 0.7V in series with its dynamic resistance of 2 \(\Omega\).
So, \(V_{D1} = 0.7V + I \cdot (2\Omega)\) and \(V_{D3} = 0.7V + I \cdot (2\Omega)\).
Step 4: Solve for the current I.
Substitute the diode models into the KVL equation:
\[ 5 - (0.7 + 2I) - 20I - (0.7 + 2I) = 0 \]
\[ 5 - 1.4 - 2I - 20I - 2I = 0 \]
\[ 3.6 - 24I = 0 \]
\[ 24I = 3.6 \]
\[ I = \frac{3.6}{24} = 0.15 \text{ A} \]
Convert the current to milliamperes: \(I = 0.15 \text{ A} = 150 \text{ mA}\).
