Question

# Find potential drop across capacitors C2, C4 and C5 in the given circuit?

A
3 V,0.6 V,15 V
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
B
15 V,3.5 V,9 V
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
C
0.6 V,3.5 V,1.5 V
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
1.5 V,12 V,9 V
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
Open in App
Solution

## The correct option is C 0.6 V,3.5 V,1.5 VWe can solve this problem by potential method. Assigning potential to known point and identifying isolated parts in the given circuit and applying Kirchoffs junction law to it as shown in figure. For isolated system I: (x−10)×6+(x−y)×10+(x−5)×8=0 ⇒6x−60+10x−10y+8x−40=0 ⇒24x−10y−100=0 ......(1) For isolated system II: (y−5)×14+(y−x)×10+(y−0)×12=0 ⇒14y−70+10y−10x+12y=0 ⇒36y−10x−70=0 ......(2) Solving equations (1) and (2): Multiplying equation (2) by 2.4 and adding to equation (1), we get y=3.5 V Substituing y value in equation (1) 24x=100+10×(3.5) x=13524=5.6 V Using the value of x and y we can get the potential drop across capacitors C2,C4 and C5 as follows. V2=(5.6−5)=0.6 V V4=(3.5−0)=3.5 V V5=(5−3.5)=1.5 V Hence, option (c) is the correct answer. Why this question: To understanding the application of kirchhoffs junction law.

Suggest Corrections
0
Explore more