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What are the values of the currents flowing in each of the following diode circuits $\mathrm{X}$ and $\mathrm{Y}$ respectively? (Assume that the diodes are ideal)
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The correct answer is:
$4 \mathrm{~A}, 2 \mathrm{~A}$
In circuit $\mathrm{X}$, both the diodes are forward biased and hence both will conduct.
The two resistances of $4 \Omega$ each are in parallel. Their equivalent resistance is $2 \Omega$. Hence the current $\mathrm{I}=\frac{\mathrm{v}}{\mathrm{R}}=\frac{8}{2}=4 \mathrm{~A}$
In the circuit $Y$, the diode $D_1$ is forward biased but diode $D_2$ is reverse biased. Hence only diode $\mathrm{D}_1$ will conduct. The resistance is $4 \Omega$.
Hence, $\mathrm{I}=\frac{8}{4}=2 \mathrm{~A}$
The two resistances of $4 \Omega$ each are in parallel. Their equivalent resistance is $2 \Omega$. Hence the current $\mathrm{I}=\frac{\mathrm{v}}{\mathrm{R}}=\frac{8}{2}=4 \mathrm{~A}$
In the circuit $Y$, the diode $D_1$ is forward biased but diode $D_2$ is reverse biased. Hence only diode $\mathrm{D}_1$ will conduct. The resistance is $4 \Omega$.
Hence, $\mathrm{I}=\frac{8}{4}=2 \mathrm{~A}$
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