Two junction diodes, one of germanium `(Ge)` and other of silicon `(Si)` are connected as shown in fig to a battery of `12 V` and a load resistance `10 kOmega`. The germanium diode conducts at `0.3 V` and silicon diode at `0.7 V`. When current flows in the circuit, the potential of terminal `Y` will be
.

Options

A. 12V
B. 11.7V
C. 11.3V
D. 11V

Answer

B

Explanation

The junction diode of Ge conducts at `0.3V` and that of Si at `0.70V`. As they are connected in parallel, the circuit will start conduting when potential difference across the parallel combination is `0.3V`. Here, junction diode of Si will not conduct, only junction diode of Ge will conduct. Since
applied voltage = sum of pot. drop across parallel combination and load resistance
`:. 12=0.3 +` pot. diff. across load of `10 kOmega`
`:.` pot. diff . across load of `10 kOmega=12-0.3=11.7V` (which is the pot. at Y)