Practice Problems (KEY)

Practice Problems (KEY)#

Problem 1

(a) Find \(I_S\):

\[R_{\text{eq}} = R_1 + \left(\frac{1}{R_2} + \frac{1}{R_3 + R_4 + R_5}\right)^{-1} + R_6 = 8.71\,\Omega\]
\[I_S = \frac{V_S}{R_{\text{eq}}} = \frac{5\,\text{V}}{8.71\,\Omega} = \boxed{574\,\text{mA}}\]

(b) Find the power consumed by \(R_2\) (current divider):

\[I_2 = I_S \frac{R_{2,3,4,5}}{R_2} = (574\,\text{mA})\frac{1.71\,\Omega}{2\,\Omega} = 491\,\text{mA}\]
\[P_2 = I_2^2 R_2 = (0.491\,\text{A})^2(2\,\Omega) = \boxed{482\,\text{mW}}\]

Verification using voltage divider:

\[V_2 = V_S \frac{R_{2,3,4,5}}{R_{\text{eq}}} = (5\,\text{V})\frac{1.71\,\Omega}{8.71\,\Omega} = 982\,\text{mV}\]
\[P_2 = \frac{V_2^2}{R_2} = \frac{(0.982\,\text{V})^2}{2\,\Omega} = 482\,\text{mW} \checkmark\]
Problem 2

\[R_{\text{eq}} = R_1 + R_{2,3} = 50\,\Omega + \left(\frac{1}{500\,\Omega} + \frac{1}{500\,\Omega}\right)^{-1} = \boxed{300\,\Omega}\]
\[I_S = \frac{V_S}{R_{\text{eq}}} = \frac{1.5\,\text{V}}{300\,\Omega} = \boxed{5\,\text{mA}}\]
\[P_{\text{Total}} = I_S V_S = (5\,\text{mA})(1.5\,\text{V}) = \boxed{7.5\,\text{mW}}\]
\[I_2 = \frac{R_{2,3}}{R_2} I_S = \frac{250\,\Omega}{500\,\Omega}(5\,\text{mA}) = 2.5\,\text{mA}\]
\[P_2 = I_2^2 R_2 = (2.5\,\text{mA})^2(500\,\Omega) = \boxed{3.125\,\text{mW}}\]
Problem 3

\[R_{\text{EQ}} = \left(\frac{1}{470\,\Omega} + \frac{1}{330\,\Omega} + \frac{1}{1.2\,\text{k}\Omega + 1\,\text{k}\Omega}\right)^{-1} = 182.78\,\Omega\]
\[I_S = \frac{V_S}{R_{\text{EQ}}} = \frac{15\,\text{V}}{182.78\,\Omega} = 82.10\,\text{mA}\]

No, the fuse rating is not high enough: \(82.1\,\text{mA} > 60\,\text{mA}\) — the fuse will pop.

The fuse should be rated between:

\[I_{\text{fuse,min}} = 1.1 \times 82.1\,\text{mA} = 91\,\text{mA}\]
\[I_{\text{fuse,max}} = 1.5 \times 82.1\,\text{mA} = 123\,\text{mA}\]

Choose a 100 mA circuit breaker from the parts bin.

Problem 4

\[V_{\text{Adapter}} = V_S - V_{\text{Lamp}} = 115\,\text{V} - 45\,\text{V} = 70\,\text{V}\]
\[R_{\text{Adapter}} = \frac{V_{\text{Adapter}}}{I_S} = \frac{70\,\text{V}}{2\,\text{A}} = \boxed{35\,\Omega}\]
Problems 5–15

(Solved in class — see lecture notes for full solutions.)