Assume you are given three capacitors as follows: C1 = 1 uF and 20V max. C2 = 2

Question

Assume you are given three capacitors as follows: C1 = 1 uF and 20V max. C2 = 2 uF and 50V max. C3 = 3 uF and 100V max. The maximum possible capacitance that could be attained by using the capacitors singly, or in any combination of two, or all three at once, would be (assume that the maximum voltage is not an issue): A) 1 uF (B) 2 uf C) 3uF D) 6 uF E) none of the above The minimum possible capacitance that could be attained by using the capacitors singly, or in any combination of two, or all three at once, would be (assume that the maximum voltage is not an issue): A) 1/3 uF B) 1/2 uF C) 1 uF D) 11/6 uF E) 6/11 uF Assume that all three capacitors were connected in parallel. The maximum voltage that could be safely applied to the parallel combination (without causing dielectric breakdown) would be just less than: A) 20v B) 50V C) Tov D) 100V E) 170v Assume that all three capacitors were connected in series The maximum voltage that could be sai applied to the series combination (without causing dielectric breakdown) would be just less than: A) 20.00V B) 36.67V C) 50.00 D) 63.33V E) 100.00V When a current passes through a conductor, the actual charge carriers A. always move in the direction of the current. B. always move from higher to lower potential. C. always move in the same direction D. always have the same sign E. none of the above

Explanation / Answer

1. Maximum Capacitance can be attained by combining these capacitors in parallel
Then Cmax = 1+2+3 = 6uF
So option D.

2. Minimum Capacitance can be attained by combining these capacitors in series.
Then Cmin = 1/ (1/1+1/2+1/3) = 6/11 uF

3. 20+50+100 = 170 V

4. Less than 20 V

5. B. moves from high to low potential.

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