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In the circuit shown assume the diode to be ideal. When $V_i$ increases from $-2 \mathrm{~V}$ to $6 \mathrm{~V}$, the change in current is (in $\mathrm{mA})$
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Verified Answer
The correct answer is:
$20$
$20$
$$
l_{\text {initial }}=0 \text { for } V_i \leq+1 \mathrm{~V}
$$
This diode conduct only beyond $V_i=+1 \mathrm{~V}$
$$
I_{\text {final }}=\frac{5}{250}=0.02 \mathrm{~A}
$$
So, change in $I=0.02 \mathrm{~A}=20 \mathrm{~mA}$
l_{\text {initial }}=0 \text { for } V_i \leq+1 \mathrm{~V}
$$
This diode conduct only beyond $V_i=+1 \mathrm{~V}$
$$
I_{\text {final }}=\frac{5}{250}=0.02 \mathrm{~A}
$$
So, change in $I=0.02 \mathrm{~A}=20 \mathrm{~mA}$
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