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Find the current through $6\Omega$ resistor using superposition theorem.
1 Answer
written 7.8 years ago by |
$I_{6\Omega}^1=\dfrac{I_1(5)}{6+5}.......(1)$
$(5^{-1}+6^{-1})^{-1}=2.727 \Omega$
$(2+2.727)=4.727 \Omega$
$$I_1=\dfrac{4(10)}{[10+4.727]}=2.716 A \\ From(1) \\ I_{6\Omega}^1=\dfrac{(2.716)(5)}{11}1.234 A (\downarrow)$$
$(5^{-1}+6^{-1})^{-1}=2.727 \Omega$
$$I=\dfrac{10}{(10+4.727)}=0.679 A \\ I_{6\Omega}^{11}=\dfrac{I(5)}{(6+5)}=0.3086A(\uparrow)$$
$$I_{6\Omega}^{111}=\dfrac{3(3.529)}{6+3.259}=1.111A(\downarrow)$$
Source | Current |
---|---|
4 A current source | I6Ω1=1.234 A(↓) |
10 V … |