The circuit contains two diodes,each with a f | vedclass

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(A) In the given circuit,diode $D_{1}$ is forward-biased because its p-terminal is connected to the positive terminal of the battery. Diode $D_{2}$ is

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$20$

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(A) In the given circuit,diode $D_{1}$ is forward-biased because its p-terminal is connected to the positive terminal of the battery. Diode $D_{2}$ is reverse-biased because its n-terminal is connected to the positive terminal of the battery.Since $D_{2}$ is reverse-biased,it acts as an open circuit,and no current flows through the branch containing $D_{2}$ and the $100\, \Omega$ resistor.The circuit simplifies to a series combination of the battery $(6\, V)$,the forward resistance of $D_{1}$ $(50\, \Omega)$,the resistor $R_{1}$ $(130\, \Omega)$,and the resistor $R_{3}$ $(120\, \Omega)$.The total resistance of the circuit is $R_{total} = 50\, \Omega + 130\, \Omega + 120\, \Omega = 300\, \Omega$.Using Ohm's law,the current $i$ flowing through the circuit is:$i = \frac{V}{R_{total}} = \frac{6\, V}{300\, \Omega} = 0.02\, A$.Converting this to milliamperes $(mA)$:$i = 0.02 	imes 1000\, mA = 20\, mA$.Thus,the current through the $120\, \Omega$ resistance is $20\, mA$.

The circuit contains two diodes,each with a forward resistance of $50\, \Omega$ and an infinite reverse resistance. If the battery voltage is $6\, V$,the current through the $120\, \Omega$ resistance is $mA$.

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