Discussion
For the following diode circuit, the voltage drop across each diode is 0.7 V. The current flowing through the resistor R is .............
Answer: C
The current is +8 mA.
The three diodes are in series and forward biased, so together they drop 3 × 0.7 = 2.1 V. The rest of the supply appears across R:
\[ \begin{aligned} I &= \frac{10.1 - 3(0.7)}{1.0\ \text{k}\Omega} \\ &= \frac{8.0\ \text{V}}{1.0\ \text{k}\Omega} = 8\ \text{mA} \end{aligned} \]
flowing from the positive supply down through R and the diodes, so positive in the direction shown. 10.1 mA ignores the diode drops, and the negative values would mean current flowing against the forward direction of the diodes, which they would block.
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