In the figure shown, if the diode forward vol | vedclass

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(B) Given current $I = 0.2 \, mA = 0.2 	imes 10^{-3} \, A$.Voltage drop across the diode $V_d = 0.2 \, V$.Resistors $R_1 = R_2 = 5 \, k\Omega = 5 	i

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$2.2$

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(B) Given current $I = 0.2 \, mA = 0.2 	imes 10^{-3} \, A$.Voltage drop across the diode $V_d = 0.2 \, V$.Resistors $R_1 = R_2 = 5 \, k\Omega = 5 	imes 10^3 \, \Omega$.The circuit consists of a resistor $R_1$, a diode, and a resistor $R_2$ in series.The voltage drop across resistor $R_1$ is $V_{R1} = I 	imes R_1 = (0.2 	imes 10^{-3} \, A) 	imes (5 	imes 10^3 \, \Omega) = 1 \, V$.The voltage drop across resistor $R_2$ is $V_{R2} = I 	imes R_2 = (0.2 	imes 10^{-3} \, A) 	imes (5 	imes 10^3 \, \Omega) = 1 \, V$.The total voltage difference between $A$ and $B$ is the sum of the voltage drops across the components: $V_{AB} = V_{R1} + V_d + V_{R2} = 1 \, V + 0.2 \, V + 1 \, V = 2.2 \, V$.

In the figure shown, if the diode forward voltage drop is $0.2 \, V$, the voltage difference between $A$ and $B$ is (in $V$)

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