A silicon diod is in series with a 1 kilo ohm resistor and a 5 volt battery.If the anode is connected to the positive battery terminal the cathode voltage with respect to the negative battery terminal is:
C) 5.7 v
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4.3V. The forward voltage drop across standard silicon diodes is 0.7V (same as a BJT's base-emitter drop, incidentally). 5V from the source minus 0.7V from the diode leaves 4.3V across the resistor.
Incidentally, you can now use ohm's law on the resistor to determine the current through it. Since the current through series elements will be equal, you also now know the current through the diode. This is very useful in sizing resistors for LED circuits.
The diode voltage depends on the current through it and the diode design. Generally it will be less the .9 V and greater then 0 when forward biased. So the answer to this question could be a diode drop of .3 or .7 dependent on diode design and the voltage drop across the 1K resistor.