The answer is 0.
Assume that the diode D1 is reverse biased. Since the input voltage at the anode of the diode is negative.
Here, the diode acts as an open circuit. Thus, no current flows in the circuit [tex]$(i=0)$[/tex].
Apply the Kirchhoff's voltage law.
[tex]$\begin{aligned}&-V_{1}+V_{D 1}+i(272)=0 \\&-(-3)+V_{D 1}+(0)(272)=0 \\&3+V_{D 1}=0 \\&V_{D 1}=-3 \mathrm{~V}\end{aligned}$[/tex]
Hence, the assumption of diode D1 is reverse biased is true. Since, the voltage across the diode D1 is less than 0.65 volts.
Therefore, the voltage [tex]$V_{R 1}$[/tex] is,
[tex]$\begin{aligned}V_{R \mathrm{I}} &=272 i \\&=272(0) \\&=0 \mathrm{~V}\end{aligned}$[/tex]
What is diode?
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