Chapter 8: Q12P (page 378)

Derive Eq. 8.43. [Hint: Use the method of Section 7.2.4, building the two currents up from zero to their final values.]

Short Answer

Expert verified

The equation 8.43 is derived as $W = \frac{1}{2} L_{a} l_{a} + \frac{1}{2} L_{b} l_{b} + M L_{a} l_{b}$ .

Step by step solution

Expression for the power delivered to the two loops:

Write the expression for the power delivered to the two loops.

$W = \frac{1}{2} L l^{2}$

Here, L is the inductance, and I is the current.

Derive equation 8.43 as W=12LaIa2+12LbIb2+MIaIb

Write the expression for the total energy stored in the pair of coils.

$\frac{d W}{d t} = - ε_{a} l_{a} - ε_{b} l_{b} . . . . . . . . (1)$

Where,

$ε_{a} = - L_{a} \frac{d l}{d t} - M \frac{d l_{a}}{d t} = - L_{b} \frac{d l_{b}}{d t} - M \frac{d l_{b}}{d t}$

Substitute the known values in equation (1).

$\frac{d W}{d t} = - (- L_{a} \frac{d l_{a}}{d t} - M \frac{d l_{a}}{d t}) - (- L_{a} \frac{d l}{d t} - M \frac{d l_{b}}{d t}) \frac{d W}{d t} = - (- L_{a} l_{a} \frac{d l_{a}}{d t} - M l_{a} \frac{d l_{b}}{d t}) + (L_{b} \frac{d l}{d t} + M \frac{d l_{b}}{d t}) \frac{d W}{d t} = \frac{d}{d t} (- \frac{1}{2} L_{a} l_{a}^{2} + \frac{1}{2} L_{a} l_{b}^{2} + M l_{a} l_{b})$

Integrate with respect to t.

$W = - \frac{1}{2} L_{a} l_{a}^{2} + \frac{1}{2} L_{b} l_{b}^{2} + M l_{a} l_{b}$

Therefore, equation 8.43 is derived as $W = - \frac{1}{2} L_{a} l_{a}^{2} + \frac{1}{2} L_{b} l_{b}^{2} + M l_{a} l_{b}$ .

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Derive Eq. 8.43. [Hint: Use the method of Section 7.2.4, building the two currents up from zero to their final values.]

Short Answer

Step by step solution

Expression for the power delivered to the two loops:

Derive equation 8.43 as W=12LaIa2+12LbIb2+MIaIb

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