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In the circuit shown in figure find <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/DCP_VOL_4_C25_S01_010_Q01.png" width="80%"> <br> a. the equivalent capacitance <br> b. the charge stored in each capacitor and <br> c. the potential difference across each capacitor.

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Answer Text

Solution :
a. The equivalent capacitance <br> `C=(C_1C_2)/(C_1+C_2)` <br> or `C=((2)(3))/(2+3)=1.2muF` <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/DCP_VOL_4_C25_S01_010_S01.png" width="80%"> <br> b. The charge `q` stored in each capacitor is <br> `q=CV=(1.2xx10^-6)(100)C` <br> `=120muC` <br> c. In series combination `Vprop1/C or V_1/V_2=C_2/C_1` <br> `:. V_1=(C_2/(C_1+C_2))V+((3)/(2+3))(100)=60V`<br> `and V_2=V-V_1 =100-60=40V` ltbr In parallel <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/DCP_VOL_4_C25_S01_010_S02.png" width="80%"> <br> the arrangement shown in figure is called a paralel connection. In a parallel combination the potential difference for all individual capacitors is the same and the total charge `q` is distributed in the ratio of the capacities. (as`q=CV` or `qpropC` for same potential difference). Thus, <br> `q_1/q_2=C_1/C_2` <br> or `q_1=((C_1)/(C_1+C_2))q and q_2=((C_2)/(C_1+C_2))q` <br> The parallel combination is equivalent to a single capacitor with the same total charge `q=q_1+q_2` and potential difference V. <br> Thus, <br> `q=q_1+q_2` <br> or `CV=C_1+C_2V` <br> or `C=C_1+C_2` <br> In the same way, we can show that for any number of capacitors in parallel <br> `C=C-1+C_2+C_3+..........` <br> In a parallel combination the equilavent capacitance is always greater than any individual capacitance.

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