For one approach to problems like this one, consult Interactive Solution 20.107.

Question

For one approach to problems like this one, consult Interactive Solution 20.107. Four identical capacitors are connected with a resistor in two different ways. When they are connected as in part a of the drawing, the time constant to charge up this circuit is 0.61 s. What is the time constant when they are connected with the same resistor as in part b?

For one approach to problems like this one, consult Interactive Solution 20.107. Four identical capacitors are connected with a resistor in two different ways. When they are connected as in part a of the drawing, the time constant to charge up this circuit is 0.61 s. What is the time constant when they are connected with the same resistor as in part b?

Explanation / Answer

In part a

the equivalent capacitance is 2C

in part b

it is 1/Ceq = 1/2C + 1/C + 1/C

so C eq = 0.4C

So the time constant

0.4/2*0.61 = 0.32s

the time constant when they are connected with the same resistor as in part b is 0.32 sec

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