A 2,V battery is connected across AB as shown | vedclass

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(A) $1$. When the positive terminal is connected to $A$,the diode $D_1$ is forward-biased and $D_2$ is reverse-biased. The circuit behaves as a resist

Vedclass · Accepted Answer

$0.4\,A$ and $0.2\,A$

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(A) $1$. When the positive terminal is connected to $A$,the diode $D_1$ is forward-biased and $D_2$ is reverse-biased. The circuit behaves as a resistor of $5\,\Omega$ in series with the battery.$2$. The current $I_1 = \frac{V}{R_1} = \frac{2\,V}{5\,\Omega} = 0.4\,A$.$3$. When the positive terminal is connected to $B$,the diode $D_2$ is forward-biased and $D_1$ is reverse-biased. The circuit behaves as a resistor of $10\,\Omega$ in series with the battery.$4$. The current $I_2 = \frac{V}{R_2} = \frac{2\,V}{10\,\Omega} = 0.2\,A$.$5$. Thus,the currents are $0.4\,A$ and $0.2\,A$ respectively.

$A$ $2\,V$ battery is connected across $AB$ as shown in the figure. The value of the current supplied by the battery when in one case the battery's positive terminal is connected to $A$ and in the other case when the positive terminal of the battery is connected to $B$ will respectively be:

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