Explanation Given: The following equation is given from the previous solution of exercise 55: q ( t ) = C V − C V e − t R C Where Resistance R = 200 ohms , Capacitor C = 0.02 farad and V = 12 Volts . And, time t is in seconds. Calculation: Consider the equation: q ( t ) = C V − C V e − t R C Put the values of R , C and V in the equation, we get q ( t ) = C V − C V e − t R C = 0.02 × 12 ( 1 − e − t 200 × 0.02 ) q ( t ) = 0.24 ( 1 − e − t 4 ) … ( 1 ) Now, the limiting value charge for the capacitor is: q = C V = 0...

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Chapter 10.1, Problem 58E

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We need to determine the time it takes for the charge on the capacitor plates to reach half of its limiting value.

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Chapter 10.1, Problem 57E

Chapter 10.1, Problem 59E

Chapter 10 Solutions

CALCULUS W/SAPLING ACCESS >IC<

Ch. 10.1 - Prob. 1PQ Ch. 10.1 - Prob. 2PQ Ch. 10.1 - Prob. 3PQ Ch. 10.1 - Prob. 4PQ Ch. 10.1 - Prob. 1E Ch. 10.1 - Prob. 2E Ch. 10.1 - Prob. 3E Ch. 10.1 - Prob. 4E Ch. 10.1 - Prob. 5E Ch. 10.1 - Prob. 6E

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We need to determine the time it takes for the charge on the capacitor plates to reach half of its limiting value.

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We need to determine the time it takes for the charge on the capacitor plates to reach half of its limiting value.

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Chapter 10 Solutions