Two coils having inductance of 0.2H and 2.45H are coupled and their mutual inductance is 0.14H. The coefficient of coupling is
0.2
Understanding the relationship between self-inductance, mutual inductance, and the coefficient of coupling is fundamental in the study of coupled circuits. This problem requires us to calculate the coefficient of coupling (k) given the individual inductances of two coils and their mutual inductance.
The coefficient of coupling (\(k\)) is defined by the following formula:
\[k = \frac{M}{\sqrt{L_1 L_2}}\]
Where:
Let's use the given values to calculate the coefficient of coupling.
Step 1: Calculate the product of the self-inductances (\(L_1 L_2\)).
\[L_1 L_2 = 0.2 \, \text{H} \times 2.45 \, \text{H} = 0.49 \, \text{H}^2\]
Step 2: Calculate the square root of the product of the self-inductances (\(\sqrt{L_1 L_2}\)).
\[\sqrt{L_1 L_2} = \sqrt{0.49} = 0.7 \, \text{H}\]
Step 3: Apply the coefficient of coupling formula.
\[k = \frac{M}{\sqrt{L_1 L_2}} = \frac{0.14 \, \text{H}}{0.7 \, \text{H}}\]
\[k = \frac{0.14}{0.7} = \frac{14}{70} = \frac{1}{5} = 0.2\]
The calculated coefficient of coupling \(k\) is 0.2. This value indicates that there is some magnetic coupling between the two coils, but it is not perfect coupling (which would be \(k=1\)). This type of calculation is crucial in designing and analyzing transformers, induction motors, and other magnetically coupled circuits.
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