In connection with the circuit drawn below, the value of current flowing through 2 kΩ resistor is _________ × 10⁻⁴ A.

Solution

Physics sir · 3 steps

First check that the Zener really breaks down; after that, the 2 kΩ sits across a fixed 5 V.

1. Is it in breakdown. Remove the Zener for a moment. The 1 kΩ and 2 kΩ then divide 10 V, so the 2 kΩ would get 10 × 2/3 = 6.67 V. That is more than 5 V, so the reverse-biased Zener does break down.
2. Voltage is clamped. In breakdown the Zener holds the voltage across itself at 5 V. The 2 kΩ is in parallel with it, so it also has exactly 5 V across it.

   I_(2 kΩ) = (5 V)/(2000 Ω) = 2.5 × 10⁻³ A

3. So. 2.5 × 10⁻³ A = 25 × 10⁻⁴ A, so the answer is 25. The 1 kΩ carries (10 - 5)/1000 = 50 × 10⁻⁴ A, and the other 25 × 10⁻⁴ A goes through the Zener.

The common slip: 33. Ignoring the Zener and using 10/3000 A gives about 33. The Zener clamps the 2 kΩ at 5 V, so its current is lower.

The idea to keep. For a Zener in parallel with a load, first find the load voltage without the Zener. If it exceeds V_Z, the load voltage is pinned at V_Z and Ohm's law gives the current.

Answer from the NTA final answer key. Solution by PrepWiser.