The circuit shown below contains two ideal di | vedclass

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(B) In the given circuit,diode $D_1$ is forward-biased and diode $D_2$ is reverse-biased.Since $D_2$ is reverse-biased,it acts as an open circuit and

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$0.020$

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(B) In the given circuit,diode $D_1$ is forward-biased and 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 $D_2$.The circuit effectively consists of the $6\,V$ battery,the $100\,\Omega $ resistor,the diode $D_1$ (with forward resistance $50\,\Omega $),and the $150\,\Omega $ resistor in series.The total resistance of the circuit is $R_{total} = R_{diode} + R_1 + R_{series} = 50\,\Omega + 150\,\Omega + 100\,\Omega = 300\,\Omega$.The current $I$ flowing through the $100\,\Omega $ resistor is given by Ohm's law: $I = \frac{V}{R_{total}} = \frac{6\,V}{300\,\Omega} = 0.020\,A$.

The circuit shown below contains two ideal diodes,each with a forward resistance of $50\,\Omega $. If the battery voltage is $6\,V$,the current through the $100\,\Omega $ resistance (in Amperes) is

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