9. (0 pts) Find the current $i$ in the figure below using the principle of superposition. 20 Ω W 10 Ω 40 V 6 A
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The circuit becomes a simple voltage divider. The current $i_1$ due to the 40 V source is given by: $i_1 = \frac{40}{20 + 10} = \frac{40}{30} = \frac{4}{3} A$ Show more…
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what is the current in the circuit of the figure? (figure 1)
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For the circuit shown find the current in each resistor and the current drawn from the 40 - $V$ source. Notice that the p.d. from $a$ to $b$ is $40 \mathrm{~V}$. Therefore, the p.d. across each resistor is $40 \mathrm{~V}$. Then, $$ I_{2}=\frac{40 \mathrm{~V}}{2.0 \Omega}=20 \mathrm{~A} \quad I_{5}=\frac{40 \mathrm{~V}}{5.0 \Omega}=8.0 \mathrm{~A} \quad I_{8}=\frac{40 \mathrm{~V}}{8.0 \Omega}=5.0 \mathrm{~A} $$ Because $I$ splits into three currents: $$ I=I_{2}+I_{5}+I_{8}=20 \mathrm{~A}+8.0 \mathrm{~A}+5.0 \mathrm{~A}=33 \mathrm{~A} $$
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