The cut-off voltage of the diodes (shown in f | vedclass

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(B) In the given circuit,diode $D_1$ is forward-biased,while diode $D_2$ is reverse-biased. Therefore,no current flows through the branch containing $

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$4$

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(B) In the given circuit,diode $D_1$ is forward-biased,while diode $D_2$ is reverse-biased. Therefore,no current flows through the branch containing $D_2$.The circuit simplifies to a series combination of the $1 \,V$ battery,the $60 \,\Omega$ resistor,the forward-biased diode $D_1$ (with a voltage drop of $0.6 \,V$),and the $40 \,\Omega$ resistor.Applying Kirchhoff's Voltage Law $(KVL)$ in the loop:$1 - I(60) - 0.6 - I(40) = 0$Simplifying the equation:$0.4 - I(100) = 0$$I(100) = 0.4$$I = \frac{0.4}{100} \,A$$I = 0.004 \,A = 4 \,mA$Thus,the current through the $40 \,\Omega$ resistor is $4 \,mA$.

The cut-off voltage of the diodes (shown in figure) in forward bias is $0.6 \,V$. The current through the resistor of $40 \,\Omega$ is ........... $mA$.

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