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14.Semiconductor Electronics
medium
$R=............ \;k \Omega$
A
$3$
B
$2$
C
$5$
D
$4$
(AIIMS-2019)
Solution
The current across the circuit is,
$i=\frac{30}{R+1} mA$
The voltage across the resistance $1 k \Omega$,
$V_{z}=i(1 k \Omega)$
$6=\frac{30}{R+1} \operatorname{mA}(1 k \Omega)$
$30=6 R+6$
$R=4 k \Omega$
Standard 12
Physics
