Assuming an ideal diode,find the current thro | vedclass

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(A) In the given circuit,the top branch contains a diode in forward bias,and the middle branch contains a Zener diode in reverse bias (as the cathode

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$2 \ A$

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(A) In the given circuit,the top branch contains a diode in forward bias,and the middle branch contains a Zener diode in reverse bias (as the cathode is connected to the positive terminal of the battery). $1$. The diode in the top branch is in forward bias,so it acts as a short circuit (ideal diode). The resistance in this branch is $2 \Omega$. $2$. The Zener diode in the middle branch is in reverse bias,so it acts as an open circuit. No current flows through this branch. $3$. The total resistance of the circuit is the resistance of the top branch $(2 \Omega)$ in parallel with the middle branch (open circuit,so infinite resistance),which is $2 \Omega$. This is then in series with the $1 \Omega$ internal resistance. $4$. Total resistance $R_{eq} = 2 \Omega + 1 \Omega = 3 \Omega$. $5$. The current through the $1 \Omega$ resistor is the total current supplied by the $6 \ V$ battery: $I = \frac{V}{R_{eq}} = \frac{6 \ V}{3 \Omega} = 2 \ A$.

Assuming an ideal diode,find the current through the $1 \Omega$ resistance in the circuit shown in the figure.

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