In the circuit shown below, maximum zener diode current will be _________mA.

$l=\frac{120-60}{4000}=0.015A$

Thus l₂ = l - L_L

= 0.015 – 0.006

= 0.009 = 9mA

<p><math><mrow><mi>l</mi><mo>=</mo><mfrac><mrow><mn>1</mn><mn>2</mn><mn>0</mn><mo>−</mo><mn>6</mn><mn>0</mn></mrow><mrow><mn>4</mn><mn>0</mn><mn>0</mn><mn>0</mn></mrow></mfrac><mo>=</mo><mn>0</mn><mo>.</mo><mn>0</mn><mn>1</mn><mn>5</mn><mi>A</mi></mrow></math></p><p>Thus l₂ = l - L_L</p><p>= 0.015 – 0.006</p><p>= 0.009 = 9mA</p><div class="photo-widget-full"><div class="figure"><picture><source srcset="" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1754974517phppN54Sm.jpeg" alt="" width="251" height="79" loading="lazy"></picture></div></div>

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