Part C Find the charge on the 75 nF capacitor. Express your answers in microcoul

Question

Part C

Find the charge on the 75 nF capacitor.

Express your answers in microcoulombs to one decimal place.

Part D

Find the total energy stored in the network.

Express your answer in millijoules to three significant figures.

Part E

Find the energy stored in the 35 nF capacitor.

Express your answers in millijoules to two decimal places.

Part F

Find the energy stored in the 75 nF capacitor.

Express your answers in millijoules to two decimal places.

Part G

Find the potential difference across the 35 nF capacitor.

Express your answers in volts as an integer.

Part H

Find the potential difference across the 75 nF capacitor.

Express your answers in volts as an integer.

Problem 18.56 For the capacitor network shown in (Figure 1), the potential difference across ab is 240 V Figure 1 of 1 35 nF 75 nF Part A Find the total charge stored in this network Express your answer in microcoulombs to three significant figures Submit My Answers Give Up incorrect; Try Again; 4 attempts remaining Part B Find the charge on the 35 nF capacitor. Express your answers in microcoulombs to one decimal place

Explanation / Answer

Left plate of both capacitors is connected to end, a and right plate of both is connected to end, b.
G and H) Hence potential difference across both capacitors is same as potial difference between a and b.
   V(75) = V(35) = 240 V
Charge on a capacitor of capacitance, C
Q = CV where V is potantial difference across capacitor.
B) Hence Q(35) = 35x10-9 x 240 = 8.4 x 10-6 C = 8.4 micro C

C) Q(75) = 75x10-9 x 240 = 8.4 x 10-6 C = 18.0 micro C

A) Charge on network = Q(35) + Q(75) = 26.4 micro C

Energy stored in a Capacitor, U = 1/2 (C V2)

E) hence U(35) = 1/2 ( 35x10-9 x 2402 ) = 1.008 x 10-3 J = 1.01 mJ
F) U(75) = 1/2 ( 75x10-9 x 2402 ) = 2.16 x 10-3 J = 2.16 mJ
D) Total energy = U(35) + U(75) = 3.17 mJ

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