Answer to Question #320347 in Physics for percy

Question

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!

Answer 1

3.

$$C=C_1+C_2+C_3=5+10+15=30\:\rm \mu F$$

$$q_1=C_1V=5*10^{-6}*240=1.2*10^{-3}\:\rm C$$

$$q_2=C_2V=10*10^{-6}*240=2.4*10^{-3}\:\rm C$$

$$q_3=C_3V=15*10^{-6}*240=3.6*10^{-3}\:\rm C$$

4.

$$\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$$ $$\frac{1}{C}=\frac{1}{5\:\rm \mu F}+\frac{1}{10\:\rm \mu F}+\frac{1}{15\:\rm \mu F}=0.367$$ $$C=2.73\:\rm \mu F$$

$$q_1=q_2=q_3=CV\\ =2.73*10^{-6}*400=1.09*10^{-3}\rm C$$

$$V_1=\frac{q_1}{C_1}=\frac{1.09*10^{-3}}{5*10^{-6}}=218\rm \: V$$

$$V_2=\frac{q_2}{C_2}=\frac{1.09*10^{-3}}{10*10^{-6}}=109\rm \: V$$

$$V_3=\frac{q_3}{C_3}=\frac{1.09*10^{-3}}{15*10^{-6}}=73\rm \: V$$