Step 1: Recall $Y_{\text{bus}$ properties.}
- Off-diagonal entry $Y_{ij}$ corresponds to $-Y_{line}(i,j)$.
- Diagonal entry $Y_{ii}$ is the sum of admittances connected to bus $i$, i.e.
\[
Y_{ii} = \sum_{j\neq i} Y_{line}(i,j) + Y_{\text{shunt},i}.
\]
Step 2: Check line admittances from off-diagonal terms.
From the matrix:
- Between bus 1 and bus 2: $Y_{12} = j10 \;\Rightarrow\; Y_{line}(1,2)= -j10$.
- Between bus 1 and bus 3: $Y_{13} = j5 \;\Rightarrow\; Y_{line}(1,3)=-j5$.
- Between bus 2 and bus 3: $Y_{23} = j4 \;\Rightarrow\; Y_{line}(2,3)=-j4$.
Thus, three lines exist: 1–2, 1–3, 2–3.
Step 3: Check diagonal entries.
- For bus 1:
\[
Y_{11} = -j15, \text{expected from lines } = -(j10+j5) = -j15.
\]
So no shunt element at bus 1.
- For bus 2:
\[
Y_{22} = -j13.5, \text{expected from lines } = -(j10+j4) = -j14.
\]
Difference: $0.5j$, indicating a shunt capacitor at bus 2.
- For bus 3:
\[
Y_{33} = -j8, \text{expected from lines } = -(j5+j4) = -j9.
\]
Difference: $j1$, indicating a shunt capacitor at bus 3.
Step 4: Interpret results.
- All three lines (1–2, 1–3, 2–3) have line charging capacitances.
- Shunt capacitances are present at bus 2 and bus 3.
Step 5: Evaluate options.
- (A) True: All three lines have finite capacitances.
- (B) False: Not only line 2–3, all lines exist. This can NOT be true.
- (C) False: Shunt capacitor is not at bus 1, but at buses 2 and 3.
- (D) False: Shunt capacitor is not only at bus 3, but also at bus 2.
% Final Answer
\[
\boxed{\text{Option (B)}}
\]
In the circuit, \( I_{\text{DC}} \) is an ideal current source, the transistors \( M_1 \), \( M_2 \) are assumed to be biased in saturation wherein \( V_{\text{in}} \) is the input signal and \( V_{\text{DC}} \) is the fixed DC voltage. Both transistors have a small signal resistance of \( R_{ds} \) and transconductance of \( g_m \). The small signal output impedance of the circuit is:

Assuming ideal op-amps, the circuit represents:

Selected data points of the step response of a stable first-order linear time-invariant (LTI) system are given below. The closest value of the time constant (in seconds) of the system is:
\[ \begin{array}{|c|c|} \hline \textbf{Time (sec)} & \textbf{Output} \\ \hline 0.6 & 0.78 \\ 1.6 & 2.8 \\ 2.6 & 2.98 \\ 10 & 3 \\ \infty & 3 \\ \hline \end{array} \]