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(D) From the circuit diagram,when the positive terminal of the $5 \, V$ battery is connected to point $X$,diode $D_{1}$ is forward-biased,while diode

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$0.43$

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(D) From the circuit diagram,when the positive terminal of the $5 \, V$ battery is connected to point $X$,diode $D_{1}$ is forward-biased,while diode $D_{2}$ is reverse-biased.Since $D_{2}$ is reverse-biased,it acts as an open circuit and no current flows through the branch containing $D_{2}$.Diode $D_{1}$ is forward-biased and acts as a silicon diode with a potential drop of $0.7 \, V$.The current $I$ supplied by the battery flows through the branch containing $D_{1}$ and the $10 \, \Omega$ resistor.Using Kirchhoff's voltage law in the loop:$5 \, V - V_{D1} - I 	imes 10 \, \Omega = 0$$5 - 0.7 = I 	imes 10$$4.3 = 10I$$I = \frac{4.3}{10} = 0.43 \, A$.

$A$ $5 \, V$ battery is connected across the points $X$ and $Y$. Assume $D_{1}$ and $D_{2}$ to be normal silicon diodes. Find the current (in $A$) supplied by the battery if the positive terminal of the battery is connected to point $X$.

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