CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

In figure is shown a system of four capacitors connected across a 10V battery. Charge that will flow from switch S when it is closed is :

309995.png

A
zero
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
20μC from a to b
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
5μC from b to a
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
5μC from a to b
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 5μC from b to a
When switch was open as shown in the figure 1, the equivalent capacitance across path PaQ is Ceq=2×32+3×106F=1.5 μF.
Similarly the equivalent capacitance across path PbQ is also Ceq=1.5 μF
When switch is closed the charge distribution across each capacitor is as shown in the figure 2.
Thus taking potential at the negative terminal of capacitor to be zero and at the switch to be V0 We can write the charges across the capacitors as follows,

For path PaQ,
Q1=(10V0)×2 μF...(i)
Q2=(10V0)×3μF...(ii)

For Path PbQ
Q1=V0×2 μF...(iii)
Q2=V0×3μF...(iv)

Using equation (iii) in equation (i) we get V0=5V
From equation (iii) and (iv) we get charges Q1&Q2 as
Q1=10μC&Q2=15μC
Thus charge flowing across switch is from point b towards point a
The net charge flowing is 15 μC10 μC = 5 μC

460168_309995_ans.png

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Series and Parallel Combination of Capacitors
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon