The current I through the 10, Omega resistor | vedclass

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(D) In the given circuit, the diode $D_{2}$ is reverse biased because its p-terminal is connected to the negative terminal of the battery, while diode

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$80$

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(D) In the given circuit, the diode $D_{2}$ is reverse biased because its p-terminal is connected to the negative terminal of the battery, while diode $D_{1}$ is forward biased.Since $D_{2}$ is reverse biased, it acts as an open circuit and no current flows through the branch containing $D_{2}$.Thus, the current flows only through the branch containing $D_{1}$ and the $10\, \Omega$ resistor.The total resistance of the circuit is $R_{eq} = 10\, \Omega + 15\, \Omega = 25\, \Omega$.The current $I$ flowing through the $10\, \Omega$ resistor is given by Ohm's law:$I = \frac{V}{R_{eq}} = \frac{2\, V}{25\, \Omega} = 0.08\, A$.Converting this to milliamperes $(mA)$:$I = 0.08 	imes 1000\, mA = 80\, mA$.

The current $I$ through the $10\, \Omega$ resistor in the given circuit is.......$mA$.

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