Answer to Question #320347 in Physics for percy

Question #320347

3. Three capacitors with capacitances 5 μF, 10 μF and 15 μF are connected in parallel. Find the capacitance of the combination. If they are connected to a battery of 240 V, find the charge in each capacitor.

 

4. Three capacitors with capacitances 4 μF, 8 μF, and 12 μF are connected in series. A voltage of 400 V is applied to the combination. (a) Find the charge in each capacitor. (b) What is the voltage across each capacitor?


with solution thank you!


1
Expert's answer
2022-03-29T12:27:31-0400

3.

C=C1+C2+C3=5+10+15=30μFC=C_1+C_2+C_3=5+10+15=30\:\rm \mu F

q1=C1V=5106240=1.2103Cq_1=C_1V=5*10^{-6}*240=1.2*10^{-3}\:\rm C

q2=C2V=10106240=2.4103Cq_2=C_2V=10*10^{-6}*240=2.4*10^{-3}\:\rm C

q3=C3V=15106240=3.6103Cq_3=C_3V=15*10^{-6}*240=3.6*10^{-3}\:\rm C

4.

1C=1C1+1C2+1C3\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}1C=15μF+110μF+115μF=0.367\frac{1}{C}=\frac{1}{5\:\rm \mu F}+\frac{1}{10\:\rm \mu F}+\frac{1}{15\:\rm \mu F}=0.367C=2.73μFC=2.73\:\rm \mu F

q1=q2=q3=CV=2.73106400=1.09103Cq_1=q_2=q_3=CV\\ =2.73*10^{-6}*400=1.09*10^{-3}\rm C

V1=q1C1=1.091035106=218VV_1=\frac{q_1}{C_1}=\frac{1.09*10^{-3}}{5*10^{-6}}=218\rm \: V

V2=q2C2=1.0910310106=109VV_2=\frac{q_2}{C_2}=\frac{1.09*10^{-3}}{10*10^{-6}}=109\rm \: V

V3=q3C3=1.0910315106=73VV_3=\frac{q_3}{C_3}=\frac{1.09*10^{-3}}{15*10^{-6}}=73\rm \: V


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