Q.Illustration 2: Compare the load lines for the two circuits of figure 2.1.6a and figure 2.1.6b.

Both circuits share the same supply, so their cutoff voltages are equal ($V_{CE(cutoff)} = V_{CC} = 10\ \text{V}$); only the saturation current differs because it depends on the collector resistor, $I_{C(sat)} = V_{CC}/R_C$.

[!ANSWER]

Figure 2.1.6a ($R_C = 1\ k\Omega$): $I_{C(sat)} = 10\ \text{mA}$, $V_{CE(cutoff)} = 10\ \text{V}$. Figure 2.1.6b ($R_C = 5\ k\Omega$): $I_{C(sat)} = 2\ \text{mA}$, $V_{CE(cutoff)} = 10\ \text{V}$.

[!TLDR]

Both load lines end at $V_{CE} = 10\ \text{V}$ on the x-axis, but the $1\ k\Omega$ circuit reaches $10\ \text{mA}$ on the y-axis while the $5\ k\Omega$ circuit reaches only $2\ \text{mA}$.

The end points of any DC load line are the cutoff point $(V_{CE} = V_{CC},\ I_C = 0)$ and the saturation point $(V_{CE} = 0,\ I_C = V_{CC}/R_C)$. The cutoff voltage is set only by the supply, so it is the same for both circuits; the saturation current is inversely proportional to the collector resistance, so a larger $R_C$ gives a smaller saturation current and a shallower load line.

Figure 2.1.6a ($R_C = 1\ k\Omega$):

Figure 2.1.6b ($R_C = 5\ k\Omega$):

Plotting both on the same axes (figure 2.1.6c), the two lines meet at the common cutoff point $10\ \text{V}$ but the smaller collector resistance of 2.1.6a produces the steeper line reaching $10\ \text{mA}$, whereas 2.1.6b is shallower, reaching only $2\ \text{mA}$.

[!ANSWER]

Figure 2.1.6a: $I_{C(sat)} = 10\ \text{mA}$, $V_{CE(cutoff)} = 10\ \text{V}$. Figure 2.1.6b: $I_{C(sat)} = 2\ \text{mA}$, $V_{CE(cutoff)} = 10\ \text{V}$. The two load lines share the cutoff point but differ in slope.