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$$
\begin{array}{l}{\text { a) Use the principle of superposition to find the voltage } v} \\ {\text { b) Find the power dissipated in the } 10 \Omega \text { resistor. }}\end{array}
$$

Case (1) Use 4A

$$
I_{1}=\frac{4 * 12}{12+2+10}=2.77 A
$$

$$
I_{4 A}=\frac{2.77*5}{5+10}=0.9233 A
$$

Case (2) Use 110 V

$$
I=\frac{V}{R}=\frac{110}{5+5.83}=10.15 \mathrm{A}
$$

$$
I_{110 v}=\frac{10.15 * 14}{14+10}=5.9233 \mathrm{A}
$$

$$
I_{10 \Omega}=I_{110v} -I_{ 4 A}
$$

$$
I_{10 \Omega}=5.9233-0.9233
$$

$$
I_{10 \Omega}=5 A
$$

$$
V=IR=5 * 10=50 v
$$

$$
P=I^{2} R=(5)^{2} * 10
$$

$$
P=250 \mathrm{w}
$$

$$
\text { dissipated }
$$

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