电路相量法基础相关问题解答

27、28都在用KCL；27、I1→ = I→ - I2→= | I - I2 | = √( 10^2 - 6^2 ) = 8A；28、I→ = I1→ + I2→ = I1 - I2 = 4 - 3 = 1A；因为I1、I2反相，向量和就是代数和。32、选CIs = 2/√2 = √2 A；IR = U/R = 100/100 = 1 A；Ic =√[ (√2)^2 - 1^2 ] = 1 A；故 1/ωc =R = 100Ω，ω = 1/(100c) = 1/[ 100 * 10 * 10^(-6) ] = 10^3；ic超前is角度φ = arccos(√2/2) = 45°ic =√2Ic cos(ωt + 90° +45° ) =√2cos( 10^3t + 135° )33、选B 。Us = 20/√2 = 10√2，UR = 10 * 1 = 10v；Uc =√[ (10√2)^2 - 10^2 ] = 10v；故 1/ωc = R = 10 Ω，ω = 1/(10c) = 1/[ 10 * 10 * 10^(-6) ] = 10^4；对照答案选项，无需再计算角度，uc = 10√2sin( 10^4t -π/2 )；