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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

$${R}=\mathrm{3}{k}\Omega+\mathrm{1}{k}\Omega+\mathrm{2}{k}\Omega=\mathrm{6}{k}\Omega=\mathrm{6000}\Omega \\ $$$${V}={IR} \\ $$$${I}=\frac{{V}}{{R}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:{I}=\frac{\mathrm{24}}{\mathrm{6000}}=\mathrm{4}×\mathrm{10}^{−\mathrm{3}} {A} \\ $$$${I}={I}_{\mathrm{1}} +{I}_{\mathrm{2}} \\ $$$$\mathrm{4}×\mathrm{10}^{−\mathrm{3}} {A}={I}_{\mathrm{1}} +{I}_{\mathrm{2}} \:\:\:...\left({i}\right) \\ $$$${loop}\:{I} \\ $$$$\mathrm{24}−\mathrm{2000}{I}−{v}_{{a}} +{v}_{{b}} =\mathrm{0} \\ $$$$\mathrm{24}−\mathrm{2000}\left({I}_{\mathrm{1}} +{I}_{\mathrm{2}} \right)−{v}_{{a}} +{v}_{{b}} =\mathrm{0} \\ $$$${v}_{{a}} −{v}_{{b}} =\mathrm{24}−\mathrm{2000}{I}_{\mathrm{1}} −\mathrm{2000}{I}_{\mathrm{2}} \:\:\:\:\:\:...\left({ii}\right) \\ $$$${loop}\:{II} \\ $$$${v}_{{a}} −\mathrm{1000}{I}_{\mathrm{1}} −\mathrm{3000}{I}_{\mathrm{1}} −{v}_{{b}} =\mathrm{0} \\ $$$${v}_{{a}} −{v}_{{b}} =\mathrm{4000}{I}_{\mathrm{1}} \:\:\:\:\:.....\left({iii}\right) \\ $$$${solve}\:{simultaneous}\:\left({ii}\right)\&\left({iii}\right) \\ $$$${v}_{{a}} −{v}_{{b}} =\mathrm{24}−\mathrm{2000}{I}_{\mathrm{1}} −\mathrm{2000}{I}_{\mathrm{2}} \\ $$$${v}_{{a}} −{v}_{{b}} =\mathrm{4000}{I}_{\mathrm{1}} \:\: \\ $$$${I}_{\mathrm{1}} =\mathrm{4}×\mathrm{10}^{−\mathrm{3}} {A}\:\:\:\:\:{I}_{\mathrm{2}} =\mathrm{0} \\ $$$${v}_{{a}} −{v}_{{b}} =\mathrm{4000}×\mathrm{4}×\mathrm{10}^{−\mathrm{3}} {A}\:\:=\mathrm{16}{V} \\ $$$${v}_{{a}} −{v}_{{b}} =\mathrm{16}{V} \\ $$$$ \\ $$

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

thanks mr spillover

$$\mathrm{thanks}\:\mathrm{mr}\:\mathrm{spillover} \\ $$

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

V_(a−b) =(4/6)×24=16 V

$${V}_{{a}−{b}} =\frac{\mathrm{4}}{\mathrm{6}}×\mathrm{24}=\mathrm{16}\:{V} \\ $$

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

thanks mr W

$$\mathrm{thanks}\:\mathrm{mr}\:\mathrm{W} \\ $$

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