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Find the value of current flowing through the $5\Omega$ resistance using superposition theorem.
written 9.5 years ago by gravatar for teamques10 teamques10 ★ 70k •   modified 5.7 years ago

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basic electrical engineering
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written 9.5 years ago by gravatar for teamques10 teamques10 ★ 70k •   modified 9.5 years ago

→ Consider 24V supply

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$10Ω||10Ω=5Ω$

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I15Ω=2.4A(→)

→ Consider 2A current source

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Apply KVL at X

$\dfrac{V_x-0}{5}-2+\dfrac{V_x-0}{10}+\dfrac{V_x-0}{10}=0 \\ V_x[5^{-1}+10^{-1}+10^{-1}]=2 \\ V_x=5V \\ I_{5\Omega}^{11}=\dfrac{(V_x-0)}{5} \\ I_{5\Omega}^{11}=1A(\leftarrow)$

→ Consider 10V supply

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$5Ω||10Ω=3.333Ω$

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$I=\dfrac{10}{13.333}=0.75A \\ I=0.75A \\ I1115Ω=\dfrac{I×10}{(5+10)} \text{(By current Division Rule)} \\ I1115Ω=0.5A(→)$

Source Current
24 V Supply I15Ω=2.4A(→)
2A Source $I_{5\Omega}^{11}=1A(\leftarrow)$
10V Supply I1115Ω=0.5A(→)

$\boxed{I5Ω=0.9A(→)}$
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