Two coils have inductances of 8 mH and 18 mH and a coefficient of coupling of 0.5. If the two coils are connected in series aiding, the total inductance will be
A. 32 mH
B. 38 mH
C. 40 mH
D. 48 mH
Answer: Option B
A. 32 mH
B. 38 mH
C. 40 mH
D. 48 mH
Answer: Option B
A. Self-inductance
B. Mutual inductance
C. Series aiding inductance
D. Capacitance
Which circuit element(s) will oppose the change in circuit current?
A. Resistance only
B. Inductance only
C. Capacitance only
D. Inductance and capacitance
Which of the following circuit elements will oppose the change in circuit current?
A. Capacitance
B. Inductance
C. Resistance
D. All of the above
A. 2.0
B. 1.0
C. 0.5
D. Zero
To find m (mutual inductance)
m = k × √L1 L2
m = 0.5 × √8 × 18
m = 0.5 × √ 144 = 0.5 × 12
m = 6
To find total inductance
L = L1 + L2 + 2M
L = 8 + 18 + 2 ( 6)
L = 8 +18 + 12
L = 38 henry
How to find the value of m
Lt=(m×2)+L1+L2
L=L1+L2+2m, with L1=8mH, L2=18mH
m=k*sqrt(L1.L2)=0.5*sqrt(8*18)=0.5*12=6
L=8+18+2*6=38mH
m = 6
total inductance= L1 + L2 + 2M ( FOR SAME DIRECTION )
mutual inductance = K*route of L1 AND L2
How clear?
it should be 18+8+6=32