Consider the circuit shown in the figure. (Assume that C₁ = 3 μF, C₂ = 6 μF, C3 = 5 μF, and C4 = 4 μF.) C₁ C Cg 90,0 V G₁ Determine the following. (a) the total energy (in m3) stored in the system mJ (b) the energy (in m3) stored by each capacitor C₁ = mJ C₂ = mJ C3 = mJ mJ (c) Which statement is true regarding the energy of the system and the Individual capacitors? O The sum of the energies stored in the individual capacitors equals the total energy stored by the system. O The sum of the energles stored in the Individual capacitors is greater than the total energy stored by the system. O The sum of the energles stored in the Individual capacitors is less than the total energy stored by the system.
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- The figure below shows three capacitors with capacitances CA = 1.00 µF, CB = 1.60 µF, and CC = 3.80 µF connected to a 6.00-V battery. (a) What is the equivalent capacitance of the three capacitors? µF(b) What charge is stored in each of the capacitors? QA = µC QB = µC QC = µC (c) What is the potential difference across each of the capacitors? ΔVA = V ΔVB = V ΔVC = VConsider the circuit shown in the figure, with C₁ = 2.42 μF and C₂ = 7.54 µF. 2 2.00 µF 6.00 µF + 90.0 V 1 C₂ (a) Find the equivalent capacitance (in µF) of the system. UF (b) Find the charge (in μC) on each capacitor. 2.42 μF capacitor UC 6.00 μF capacitor 7.54 µF capacitor 2.00 µF capacitor 오오오 (c) Find the potential difference (in V) on each capacitor. 2.42 μF capacitor V 6.00 μF capacitor V 7.54 µF capacitor V 2.00 uF capacitor V (d) Find the total energy (in m3) stored by the group. mJThe circuit in the figure below contains a 9.00 V battery and four capacitors. The two capacitors on the left and right both have same capacitance of C, = 8.40 µF. The capacitors in the top two branches have capacitances of 6.00 µF and C, = 2.40 µF. 6.00 µF C 9.00 V (a) What is the equivalent capacitance (in uF) of all the capacitors in the entire circuit? (b) What is the charge (in µC) stored by each capacitor? right 8.40 µF capacitor left 8.40 µF capacitor 2.40 µF capacitor 6.00 µF capacitor (c) What is the potential difference (in V) across each capacitor? (Enter the magnitudes.) right 8.40 µF capacitor V left 8.40 µF capacitor 2.40 µF capacitor V 6.00 µF capacitor V
- Consider the circuit shown in the figure, with C₁ = 4.62 μF and C₂ = 7.44 μF. 2.00 µF 6.00 µF + 90.0 V C₂ (a) Find the equivalent capacitance (in µF) of the system. μF (b) Find the charge (in µC) on each capacitor. 4.62 μF capacitor 6.00 μF capacitor 7.44 μF capacitor 2.00 μF capacitor 9999 μC μC μC (c) Find the potential difference (in V) on each capacitor. 4.62 µF capacitor V 6.00 μF capacitor V 7.44 μF capacitor V 2.00 μF capacitor V (d) Find the total energy (in mJ) stored by the group. mJFind the following. (In the figure, use C, = 31.00 µF and C, = 25.00 µF.) 6.00 µF C2 µF CµF 9.00 V (a) the equivalent capacitance of the capacitors in the figure above uF (b) the charge on each capacitor on the right 31.00-µF capacitor on the left 31.00-µF capacitor μC on the 25.00-µF capacitor on the 6.00-µF capacitorFour capacitors are connected as shown in the figure below. (C = 18.0 μF.) C 3.00 με th a 6.00 με |20.0 με HH i (a) Find the equivalent capacitance between points a and b. μF (b) Calculate the charge on each capacitor, taking AV ab 20.0 μF capacitor 6.00 μF capacitor 3.00 µF capacitor capacitor C 9999 HC HC HC - = 20.0 V.
- Consider the circuit shown in the figure, with C₁ = 6.92 µF and C₂ = 6.84 µF. 2.00 µF 6.00 µF + 90.0 V (a) Find the equivalent capacitance (in μF) μF (b) Find the charge (in µC) on each capacitor. 6.92 μF capacitor 6.00 μF capacitor 6.84 μF capacitor 2.00 μF capacitor mJ 9999 (c) Find the potential difference (in V) on each capacitor. 6.92 μF capacitor V 6.00 μF capacitor V 6.84 μF capacitor 2.00 μF capacitor the system. V (d) Find the total energy (in mJ) stored by the group.Consider the circuit shown in the figure, with C₁ = 5.02 μF and C₂ = 6.64 μF. 2.00 µF 6.00 uF + 90.0 V C₂ (a) Find the equivalent capacitance (in µF) of the system. 4.27 μF (b) Find the charge (in µC) on each capacitor. 5.02 µF capacitor HC 6.00 μF capacitor με 6.64 μF capacitor 2.00 μF capacitor με μC (c) Find the potential difference (in V) on each capacitor. 5.02 μF capacitor 6.00 μF capacitor 6.64 μF capacitor 2.00 μF capacitor V (d) Find the total energy (in mJ) stored by the group. mJTwo capacitors are connected in series between the terminals of a 40.0-V battery. If their capacitances are 42.5 μF and 48.5 μF, determine the following. (a) the equivalent capacitance of the system μF (b) the magnitude of charge stored on each plate of either capacitor C (c) the voltage across the 42.5 μF capacitor (Give your answer to at least one decimal place.) V (d) the voltage across the 48.5 μF capacitor V
- Consider the system of capacitors shown in the figure below (C₁ = 3.00 μF, C₂ = 5.00 μF). 2.00 μF 6.00μF 90.0 V C₂ (a) Find the equivalent capacitance of the system. 3.33 X Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. µF (b) Find the charge on each capacitor. μC (on C₁) μC (on C₂) 180 ✔ μC (on the 6.00 μF capacitor) X 120 Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. µC (on the 2.00 μF capacitor) (c) Find the potential difference across each capacitor. 60 ✔V (across (₁) 30 X Your response differs from the correct answer by more than 10%. Double check your calculations. V (across (₂) 30 ✓ V (across the 6.00 μF capacitor) 60 X…Find the following. (In the figure use C1 = 37.80 µF and C2 = 31.80 µF.) (a) the equivalent capacitance of the capacitors in the figure above µF(b) the charge on each capacitor on the right 37.80 µF capacitor µC on the left 37.80 µF capacitor µC on the 31.80 µF capacitor µC on the 6.00 µF capacitor µC (c) the potential difference across each capacitor on the right 37.80 µF capacitor V on the left 37.80 µF capacitor V on the 31.80 µF capacitor V on the 6.00 µF capacitor VConsider the circuit shown in the figure, with C, , = 5.62 µF and C, = 7.44 µF. 6.00 µF 2.00 µF C2 + 90.0 V (a) Find the equivalent capacitance (in µF) of the system. (b) Find the charge (in µC) on each capacitor. 5.62 µF capacitor 6.00 µF capacitor 7.44 µF capacitor 2.00 µF capacitor HC (c) Find the potential difference (in V) on each capacitor. 5.62 µF capacitor V 6.00 µF capacitor V 7.44 µF capacitor V 2.00 µF capacitor V (d) Find the total energy (in mJ) stored by the group. m)