모두CBT
출제 25-1/24-3/23-1/22-2/18-2/16-1그림과 같은 브리지 회로의 평형조건은?R1LR3C2R2C1G
출제 25-1/24-3/23-1/22-2/18-2/16-1
그림과 같은 브리지 회로의 평형조건은?R1LR3C2R2C1G
R1C2=R2C1,L=R2R3C1R_1C_2=R_2C_1, L=R_2R_3C_1R1C2=R2C1,L=R2R3C1
R1C1=R2C2,R2R3=C1=LR_1C_1=R_2C_2, R_2R_3=C_1=LR1C1=R2C2,R2R3=C1=L
R1C2=R2C1,R2R3=C1LR_1C_2=R_2C_1, R_2R_3=C_1LR1C2=R2C1,R2R3=C1L
R1C1=R2C2,R2R3=C1LR_1C_1=R_2C_2, R_2R_3=C_1LR1C1=R2C2,R2R3=C1L
브리지 4변은 Z1=R1+jωLZ_1=R_1+j\omega LZ1=R1+jωL, Z2=R3+1jωC2Z_2=R_3+\dfrac{1}{j\omega C_2}Z2=R3+jωC21, Z3=R2Z_3=R_2Z3=R2, Z4=1jωC1Z_4=\dfrac{1}{j\omega C_1}Z4=jωC11이고 평형조건은 마주 보는 변의 곱이 같을 것, 즉 Z1Z4=Z2Z3Z_1Z_4=Z_2Z_3Z1Z4=Z2Z3이다.좌변 =R1jωC1+LC1=\dfrac{R_1}{j\omega C_1}+\dfrac{L}{C_1}=jωC1R1+C1L, 우변 =R2R3+R2jωC2=R_2R_3+\dfrac{R_2}{j\omega C_2}=R2R3+jωC2R2.실수부에서 LC1=R2R3\dfrac{L}{C_1}=R_2R_3C1L=R2R3 → L=R2R3C1L=R_2R_3C_1L=R2R3C1, 허수부에서 R1C1=R2C2\dfrac{R_1}{C_1}=\dfrac{R_2}{C_2}C1R1=C2R2 → R1C2=R2C1R_1C_2=R_2C_1R1C2=R2C1이므로 ① R1C2=R2C1, L=R2R3C1R_1C_2=R_2C_1,\ L=R_2R_3C_1R1C2=R2C1, L=R2R3C1이 정답이다.