In which of the following circuits the diode is reverse biased?

To determine which diode is reverse biased, we need to analyze the potential difference across the p-n junction of the diode in each circuit.

For a standard diode symbol, the triangle represents the p-side (anode) and the vertical bar represents the n-side (cathode).
- A diode is forward biased when the potential at the p-side ($V_p$) is higher than the potential at the n-side ($V_n$), i.e., $V_p>V_n$.
- A diode is reverse biased when the potential at the p-side ($V_p$) is lower than the potential at the n-side ($V_n$), i.e., $V_p<V_n$.

Let us analyze the potentials for each case from the image:

Circuit (i):
The p-side is connected to $+2\text{V}$ and the n-side is connected to $+4\text{V}$ through a $5\text{Ω}$ resistor.
Here, $V_p=+2\text{V}$ and $V_n=+4\text{V}$.
Since $V_p<V_n$ ($2\text{V}<4\text{V}$), the diode is reverse biased.

Circuit (ii):
The p-side is connected to $0\text{V}$ and the n-side is connected to $-11\text{V}$ through a $3\text{Ω}$ resistor.
Here, $V_p=0\text{V}$ and $V_n=-11\text{V}$.
Since $V_p>V_n$ ($0\text{V}>-11\text{V}$), the diode is forward biased.

Circuit (iii):
The p-side is connected to $+2\text{V}$ and the n-side is connected to $-11\text{V}$ through a $10\text{Ω}$ resistor.
Here, $V_p=+2\text{V}$ and $V_n=-11\text{V}$.
Since $V_p>V_n$ ($2\text{V}>-11\text{V}$), the diode is forward biased.

Circuit (iv):
The p-side is connected to $+5\text{V}$ and the n-side is connected to $-11\text{V}$ through a resistor.
Here, $V_p=+5\text{V}$ and $V_n=-11\text{V}$.
Since $V_p>V_n$ ($5\text{V}>-11\text{V}$), the diode is forward biased.

Therefore, only in circuit (i) is the diode reverse biased.