14.Semiconductor Electronics
medium

नीचे दर्शाए गए परिपथ के लिए $I _{z}$ का मान होगा।

A

$0.15$

B

$0.05$

C

$0.1$

D

$25$

(JEE MAIN-2021)

Solution

$I=\frac{50}{1000}=50\, {mA}$

${R}=1000 \,\Omega$

$I=\frac{50}{2000}=25\, m A$

$I_{z}=I_{1000}-I_{2000}$

$=50-25=25\, {mA}$

Standard 12
Physics

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