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Question Number 210729 by peter frank last updated on 18/Aug/24

Answered by Spillover last updated on 18/Aug/24

Answered by Spillover last updated on 18/Aug/24

R=3kΩ+1kΩ+2kΩ=6kΩ=6000Ω V=IR I=(V/R) I=((24)/(6000))=4×10^(−3) A I=I_1 +I_2 4×10^(−3) A=I_1 +I_2 ...(i) loop I 24−2000I−v_a +v_b =0 24−2000(I_1 +I_2 )−v_a +v_b =0 v_a −v_b =24−2000I_1 −2000I_2 ...(ii) loop II v_a −1000I_1 −3000I_1 −v_b =0 v_a −v_b =4000I_1 .....(iii) solve simultaneous (ii)&(iii) v_a −v_b =24−2000I_1 −2000I_2 v_a −v_b =4000I_1 I_1 =4×10^(−3) A I_2 =0 v_a −v_b =4000×4×10^(−3) A =16V v_a −v_b =16V

R=3kΩ+1kΩ+2kΩ=6kΩ=6000ΩV=IRI=VRI=246000=4×10−3AI=I1+I24×10−3A=I1+I2...(i)loopI24−2000I−va+vb=024−2000(I1+I2)−va+vb=0va−vb=24−2000I1−2000I2...(ii)loopIIva−1000I1−3000I1−vb=0va−vb=4000I1.....(iii)solvesimultaneous(ii)&(iii)va−vb=24−2000I1−2000I2va−vb=4000I1I1=4×10−3AI2=0va−vb=4000×4×10−3A=16Vva−vb=16V

Commented by peter frank last updated on 18/Aug/24

thanks mr spillover

thanksmrspillover

Answered by mr W last updated on 18/Aug/24

V_(a−b) =(4/6)×24=16 V

Va−b=46×24=16V

Commented by peter frank last updated on 18/Aug/24

thanks mr W

thanksmrW

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