The diode shown in the circuit is a silicon diode. The potential difference between the points $A$ and $B$ will be
A$6 V$
B$0.6 V$
C$0.7 V$
D$0 V$
[RPMT 2002]
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A$6 V$
In the given condition diode is in reverse biasing so it acts as open circuit.
Hence potential difference between $A$ and $B$ is $6 \mathrm{~V}$
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