Question

Three capacitors \(\mathrm{C}_{1}=80 \mu \mathrm{F}, \mathrm{C}_{2}=14 \mu \mathrm{F}\) and \(\mathrm{C}_{3}=5 \mu \mathrm{F}\) are connected as shown in Fig. Both capacitors, \(\mathrm{C}_{1}\) and \(\mathrm{C}_{2}\), have initial charges of \(40 \mu \mathrm{C}\) and \(70 \mu \mathrm{C}\) respectively. Now, both switches are closed at the same time. What is the final charges stored in each-capacitor?

Your answer:

( A) \(Q_{1}=\frac{2960}{53} \mu C, Q_{2}=\frac{2870}{53} \mu C, Q_{3}=\frac{840}{53} \mu C\)

(O B) \(Q_{1}=80 \mu \mathrm{C}, \mathrm{Q}_{2}=60 \mu \mathrm{C}, \mathrm{Q}_{3}=30 \mu \mathrm{C}\)

0 C) \(Q_{1}=80 \mu C, Q_{2}=40 \mu C, Q_{3}=70 \mu C\)

O D) \(Q_{1}=80 \mu C, Q_{2}=14 \mu C, Q_{3}=5 \mu C\)

( E) \(Q_{1}=\frac{2960}{17} \mu C, Q_{2}=\frac{2870}{17} \mu C, Q_{3}=\frac{840}{17} \mu C\)

Answer 1

Initially, C2 is charged to 70 microCulomb => the left plate has +70 microCulomb and the right plate has -70microCulomb charge. Now when the switches are turned on, the left plate of C2 and the right plate of C3 are connected and they are isoolated ie they are not connected to anything else therefore , the initiial charge on the left plate of C2 will remain in between these plates. => Q2 + Q3 = +70microCulomb where Q2 and Q3 are final charges of C2 and C3 respectively. also, we have taken into assumption that the right plate of C3 will be positive, if this assumption is wrong, Q3 will solve out to be -ve value.

Similarly,Initially, C1 is charged to 40 microCulomb => the left plate has +40 microCulomb and the right plate has -40microCulomb charge. Now when the switches are turned on, the left plate of C1 and the left plate of C3 are connected and they are isolated ie they are not connected to anything else therefore , the initiial charge on the left plate of C1 will remain in between these plates. => Q1 - Q3 = +40microCulomb where Q1 and Q3 are final charges of C1 and C3 respectively.we have taken into assumption that the left plate of C3 will be -ve.

And, from the above equations we get, Q1 +Q2 = 110 microCulomb This is true since their _ve plates are connected together and isolated so, -Q1 -Q2 = -40 -70.

Now applying kirchoffs rule on the loop starting clockwise from left end of the circuit,

initially, we are going from +ve to -ve plate in C1. So, potential change is -V1 .then from -ve to +ve plate on C2 so ptential change is +V2 then from +ve to -ve plate of C3 so potential change is -V3 and we reach the starting point hence total potentil change should be zero => -V1 + V2 - V3 =0

V=Q/C => Q2/C2 - Q3/C3 - Q1/C1 = 0 =>(70-Q3)/14 - Q3/5 - (40+Q3)/80= ---- substituting from the initial eqns

=> (80*70 -80*Q3 -224*Q3 -14*40 -14*Q3)/1120 =0 => 318 Q3 = 5040 => Q3 = 840/53 microCulomb

=> Q1=40+840/53=2960/53 microCulomb and Q2 = 70- 840/53 =2870/53 microCulomb -->Option A