How much energy is stored in the capacitor1 in the circuit and what is voltage?
Hi, im doing a practice test and the answers are given but i have no idea how to do the questions. please help.
Four capacitors 1=8.0pF, 2=6.0pF and 3=4=5.0pF are connected to a battery. The voltage of the battery is unknown. The capacitor 1 acquires a charge of magnitude Q1=15nC on each plate when connected to the circuit.
16.How much energy is stored in the capacitor 퐶1when the circuitis connected?the answer is 1.4x10−5J
my attempt: U=(1/2)(1.5x10^-9 C / 8.0X10^-12 F) = 1.4*10^-7. I dont know why its to the exponent 10^-7 instead of 10^-5 (the answer) i must be doing something wrong.
What is the voltage supplied by the battery? the answer is 5.9kV.
i have no idea.
- oldschoolLv 71 month agoFavorite Answer
We know capacitors in series have the same charge. Capacitors in parallel add and have the same voltage and Q = CV or charge = capacitance*voltage. C1 and C both have 15nC.
E = ½CV² = ½QV = ½Q²/C = ½*(15e-9)²/8e-12 = 14µJ which is choice a). Assuming the right C is C2 = 6pF. C3 and C4 are in parallel and add: C3+C4 = 5+5=10pF. Thus there are 3 capacitors in series meaning they all have the same charge = 15nC and we know that V = Q/C:
V = 15e-9/8e-12 +15e-9/6e-12+15e-9/10e-12 = 5875 volts or choice c)
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- oubaasLv 71 month ago
C34 = C3+C4 = 5+5 = 10 pF
C12 = C1 // C2 = 8*6 / (8+6) = 48/14 = 24/7 pF
on capacitor C1
Q2 = Q1 = 15nC
Q1 = C1*V1
V1 = Q1/C1 = 15*10^-9*10^12/8 = 1.875*10^3 V
E = C/2*V^2 = 4*10^-12*3.516*10^6 = 14*10^-6 = 1.4*10^-5 joule
Ceq = C12 // C34 = 10*24/7 / (10+24/7) = 2.553*10^-12 F
V = Q1/Ceq = 15*10^-9*10^12/2.533 = 5.9*10^3 V