Explanation It is given that the mutual conduction of a electronic device is g m = ∂ i b ∂ e c . The current i b is given by i b = 50 ( e b + 5 e c ) 1.5 . Differentiate i b = 50 ( e b + 5 e c ) 1.5 partially with respect to e c and obtain g m as follows. ∂ i b ∂ e c = ∂ ∂ e c ( 50 ( e b + 5 e c ) 1.5 ) = 50 ∂ ∂ e c ( ( e b + 5 e c ) 1.5 ) = 50 ( 1.5 ( e b + 5 e c ) 0.5 ) ∂ ∂ e c ( e b + 5 e c ) = 75 ( e b + 5 e c ) 0

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Chapter 29.3, Problem 47E

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The value of gm when eb=200V and ec=−20V.

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Chapter 29.3, Problem 46E

Chapter 29.3, Problem 48E

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Chapter 29 Solutions

Basic Technical Mathematics

Ch. 29.1 - Practice Exercise If f(x, y) = 4xy2 – 3x2y, find...Ch. 29.1 - Prob. 1E Ch. 29.1 - Prob. 2E Ch. 29.1 - Prob. 3E Ch. 29.1 - Prob. 4E Ch. 29.1 - Prob. 5E Ch. 29.1 - Prob. 6E Ch. 29.1 - Prob. 7E Ch. 29.1 - Prob. 8E Ch. 29.1 - Prob. 9E

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The value of g m when e b = 200 V and e c = − 20 V .

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The value of gm when eb=200V and ec=−20V.

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Chapter 29 Solutions