Skip to content
IV. Exercises · Q14

Q.For the given capacitor configuration -- a 9 V9\ \text{V} battery connected across a network of a 8 μF8\ \mu\text{F}, a 6 μF6\ \mu\text{F}, a 2 μF2\ \mu\text{F} and a second 8 μF8\ \mu\text{F} capacitor (labelled by the nodes a, b, c, d in the figure) --

(a) find the charge on each capacitor,
(b) find the potential difference across each capacitor, and
(c) find the energy stored in each capacitor.
Puducherry TnboardTextbookSubjectiveImportance★★★★★
46% · 56/122 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Step 1. Working through the given network (an 8 μF8\ \mu\text{F}, an 8 μF8\ \mu\text{F}, a 6 μF6\ \mu\text{F} and a 2 μF2\ \mu\text{F} capacitor arranged so that the 9 V9\ \text{V} battery splits into three equal 3 V3\ \text{V} stages) shows that every one of the four capacitors ends up with exactly V=3 VV=3\ \text{V} across it.

Step 2. Applying Q=CVQ=CV to each labelled capacitor with V=3 VV=3\ \text{V}: the 8 μF8\ \mu\text{F} capacitor at node a gives Qa=8×3=24 μCQ_a=8\times3=24\ \mu\text{C}; the 6 μF6\ \mu\text{F} capacitor at node b gives Qb=6×3=18 μCQ_b=6\times3=18\ \mu\text{C}; the 2 μF2\ \mu\text{F} capacitor at node c gives Qc=2×3=6 μCQ_c=2\times3=6\ \mu\text{C}; the second 8 μF8\ \mu\text{F} capacitor at node d gives Qd=8×3=24 μCQ_d=8\times3=24\ \mu\text{C}.

Step 3. Applying U=12CV2=12×C×9U=\tfrac{1}{2}CV^2=\tfrac{1}{2}\times C\times9 to each: Ua=12×8×9=36 μJU_a=\tfrac{1}{2}\times8\times9=36\ \mu\text{J}, Ub=12×6×9=27 μJU_b=\tfrac{1}{2}\times6\times9=27\ \mu\text{J}, Uc=12×2×9=9 μJU_c=\tfrac{1}{2}\times2\times9=9\ \mu\text{J}, Ud=12×8×9=36 μJU_d=\tfrac{1}{2}\times8\times9=36\ \mu\text{J}. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.