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(A) In the given circuit,there are two parallel branches,each containing a diode in series with a $10 \ \Omega$ resistor.When the positive terminal of

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$0.2$

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(A) In the given circuit,there are two parallel branches,each containing a diode in series with a $10 \ \Omega$ resistor.When the positive terminal of the battery is connected to $A$ and the negative terminal to $B$,the top diode is forward-biased,while the bottom diode is reverse-biased.For the top branch: The diode acts as a short circuit (zero resistance). The total resistance is $R_1 = 10 \ \Omega$. The current is $I_1 = \frac{V}{R_1} = \frac{2 \ V}{10 \ \Omega} = 0.2 \ A$.For the bottom branch: The diode acts as an open circuit (infinite resistance). The current is $I_2 = 0 \ A$.The total current in the circuit is $I = I_1 + I_2 = 0.2 \ A + 0 \ A = 0.2 \ A$.

The forward resistance of the diode is zero and the reverse resistance is infinite. When the positive terminal of a $2 \ V$ battery is connected to point $A$ and the negative terminal to point $B$,what current will flow through the circuit (in $A$)?

Similar Questions

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