Practice Problems (KEY)#
Answer the following questions about the circuit below.
a. Find \(I_{S}\)#
\[
R_{3,4,5} = 3\,\Omega + 4\,\Omega + 5\,\Omega = 12\,\Omega
\]
\[
R_{2,3,4,5}
= \frac{R_{2}R_{3,4,5}}{R_{2} + R_{3,4,5}}
= \frac{(2\,\Omega)(12\,\Omega)}{2\,\Omega + 12\,\Omega}
= 1.71\,\Omega
\]
\[
R_{eq}
= R_{1} + R_{2,3,4,5} + R_{6}
= 1\,\Omega + 1.71\,\Omega + 6\,\Omega
= 8.71\,\Omega
\]
\[
I_{S}
= \frac{V_{S}}{R_{eq}}
= \frac{5\,\mathrm{V}}{8.71\,\Omega}
= 574\,\mathrm{mA}
\]
b. Find the power consumed by \(R_{2}\)#
Current divider, where \(R_{2,3,4,5}\) is the equivalent resistance of the parallel portion:
\[
I_{2}
= I_{S}\frac{R_{2,3,4,5}}{R_{2}}
= (574\,\mathrm{mA})\frac{1.71\,\Omega}{2\,\Omega}
= 491\,\mathrm{mA}
\]
\[
P_{2}
= I_{2}^{2}R_{2}
= (0.491\,\mathrm{A})^{2}(2\,\Omega)
= 482\,\mathrm{mW}
\]
Using a voltage divider:
\[
V_{2}
= V_{S}\frac{R_{2,3,4,5}}{R_{eq}}
= (5\,\mathrm{V})\frac{1.71\,\Omega}{8.71\,\Omega}
= 982\,\mathrm{mV}
\]
\[
P_{2}
= \frac{V_{2}^{2}}{R_{2}}
= \frac{(0.982\,\mathrm{V})^{2}}{2\,\Omega}
= 482\,\mathrm{mW}
\]
Find \(R_{eq}\), \(I_{S}\), \(P_{2}\), and \(P_{Total}\).
\[
R_{eq}
= R_{1} + R_{2,3}
= 50\,\Omega + \left(\frac{1}{500\,\Omega} + \frac{1}{500\,\Omega}\right)^{-1}
= 300\,\Omega
\]
\[
I_{S}
= \frac{V_{S}}{R_{eq}}
= \frac{1.5\,\mathrm{V}}{300\,\Omega}
= 5\,\mathrm{mA}
\]
\[
P_{Total}
= I_{S}V_{S}
= (5\,\mathrm{mA})(1.5\,\mathrm{V})
= 7.5\,\mathrm{mW}
\]
\[
I_{2}
= \frac{R_{2,3}}{R_{2}}I_{S}
= \frac{250\,\Omega}{500\,\Omega}(5\,\mathrm{mA})
= 2.5\,\mathrm{mA}
\]
\[
P_{2}
= I_{2}^{2}R_{2}
= (2.5\,\mathrm{mA})^{2}(500\,\Omega)
= 3.125\,\mathrm{mW}
\]
Find the equivalent resistance and determine if the fuse rating is high enough.
\[
R_{eq}
= \left(\frac{1}{50\,\Omega + 50\,\Omega} + \frac{1}{50\,\Omega}\right)^{-1}
= 33.3\,\Omega
\]
\[
I_{S}
= \frac{V_{S}}{R_{eq}}
= \frac{15\,\mathrm{V}}{33.3\,\Omega}
= 450\,\mathrm{mA}
\]
No, the fuse rating is not high enough.
Determine the appropriate resistor so the voltage across the load is 45 V.
\[
V_{Adapter}
= V_{S} - V_{Lamp}
= 115\,\mathrm{V} - 45\,\mathrm{V}
= 70\,\mathrm{V}
\]
\[
R_{Adapter}
= \frac{V_{Adapter}}{I_{S}}
= \frac{70\,\mathrm{V}}{2\,\mathrm{A}}
= 35\,\Omega
\]