Question:medium

If the load resistance decreases in a zener regulator, the series current:

Show Hint

In a working Zener regulator, think of the series resistor and the constant Zener voltage as setting a constant total current supply (\(I_S\)). The Zener diode then "absorbs" whatever current is not drawn by the load to keep the voltage constant.
Updated On: Feb 10, 2026
  • decreases.
  • stays the same.
  • increases.
  • equals the source voltage divided by the series resistance
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Examine a typical Zener diode regulator circuit. A basic Zener regulator comprises a series resistor (\(R_S\)) connected to an input voltage (\(V_{in}\)). A Zener diode is placed in parallel with the load resistor (\(R_L\)). The Zener diode's function is to ensure a stable voltage (\(V_Z\)) across the load.
Step 2: Calculate the series current (\(I_S\)). The series current passes through the series resistor \(R_S\). The voltage drop across \(R_S\) equals the input voltage minus the Zener voltage: \(V_{RS} = V_{in} - V_Z\). Using Ohm's law, the series current is derived as:\[I_S = \frac{V_{in} - V_Z}{R_S}\]
Step 3: Assess the impact of varying load resistance. Assuming the regulator functions correctly, the Zener diode maintains a constant \(V_Z\). The input voltage \(V_{in}\) and the series resistance \(R_S\) are also constant. Consequently, all variables in the equation for \(I_S\) remain constant, resulting in a constant series current \(I_S\).
Step 4: Analyze current distribution. The total series current divides into the Zener current (\(I_Z\)) and the load current (\(I_L\)), expressed as \(I_S = I_Z + I_L\). When the load resistance \(R_L\) decreases, the load current \(I_L = V_Z / R_L\) rises. Since \(I_S\) is constant, the Zener current \(I_Z\) must decrease to compensate. The regulator remains operational as long as \(I_Z\) is positive.
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