In connection with the circuit drawn below,th | vedclass

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(C) The Zener diode is connected in parallel with the $2\, k\Omega$ resistor. Since the Zener breakdown voltage is $5\, V$,the voltage across the $2\,

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$25$

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(C) The Zener diode is connected in parallel with the $2\, k\Omega$ resistor. Since the Zener breakdown voltage is $5\, V$,the voltage across the $2\, k\Omega$ resistor will be constant at $5\, V$ as long as the Zener diode is in the breakdown region.Using Ohm's law,the current $I$ flowing through the $2\, k\Omega$ resistor is given by:$I = \frac{V}{R} = \frac{5\, V}{2 	imes 10^3\, \Omega}$$I = 2.5 	imes 10^{-3}\, A$To express this in terms of $10^{-4}\, A$:$I = 25 	imes 10^{-4}\, A$Thus,the value is $25$.

In connection with the circuit drawn below,the value of current flowing through $2\, k\Omega$ resistor is ............. $\times 10^{-4} A$.

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