In the circuit given below, V(t) is the sinus | vedclass

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(D) In the positive half-cycle of the input $V(t)$, diode $D_1$ is forward-biased and $D_2$ is reverse-biased. The current flows through $R_1$ and $R$

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Has different peak values during positive and negative half-cycles

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(D) In the positive half-cycle of the input $V(t)$, diode $D_1$ is forward-biased and $D_2$ is reverse-biased. The current flows through $R_1$ and $R$. The voltage drop across $R$ is determined by the voltage divider rule: $V_{AB, pos} = V(t) \cdot \frac{R}{R + R_1}$.In the negative half-cycle of the input $V(t)$, diode $D_2$ is forward-biased and $D_1$ is reverse-biased. The current flows through $R_2$ and $R$. The voltage drop across $R$ is: $V_{AB, neg} = |V(t)| \cdot \frac{R}{R + R_2}$.Since $R_1 = 100 \ \Omega$ and $R_2 = 150 \ \Omega$ are different, the peak voltage values across $R$ during the positive and negative half-cycles will be different.Therefore, the output is not rectified, and it has different peak values during the positive and negative half-cycles.

In the circuit given below, $V(t)$ is the sinusoidal voltage source. The voltage drop $V_{AB}(t)$ across the resistance $R$ is:

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