1) A circuit is made with 128 V battery and four identical capacitors (C = 2.00mF) as shownFind the charge and voltage on each capacitori) With Switch S OPENii) With Switch S CLOSEDS.C

Answered: 1) A circuit is made with 128 V battery…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Related questions

Question

1) A circuit is made with 128 V battery and four identical capacitors (C = 2.00mF) as shown
Find the charge and voltage on each capacitor
i) With Switch S OPEN
ii) With Switch S CLOSED
S.
C

Expert Solution

Step 1

Equivalent capacitance of capacitors connected in parallel is given by $C_{net} = C_{1} + C_{2} + . . .$

Equivalent capacitance of capacitors connected in series is given by $\frac{1}{C_{net}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + . . .$

Wheatstone bridge principle states that if the ratio of $\frac{C_{1}}{C_{4}} = \frac{C_{2}}{C_{3}}$ , then the bridge is said to be balanced and the potential at points A and B is same and thus no current flows in the arm AB.

Step 2

(a)

When switch S is open-

Equivalent capacitance of upper two capacitors in series are-

$\frac{1}{C_{net}} = \frac{1}{C} + \frac{1}{C} \frac{1}{C_{net}} = \frac{2}{C} C_{net} = \frac{C}{2}$

Similarly the net capacitance of lower two capacitors is also $\frac{C}{2}$

Therefore, the equivalent capacitance of the circuit is given by-

$\begin{array}{rcl} C_{eqv .} & = & \frac{C}{2} + \frac{C}{2} \\ C_{eqv .} & = & C \end{array}$

So total charge flowing in the circuit is-
$Q = C_{eqv .} V Q = C \times 128 Q = 128 C Q = 128 \times 2 \times 10^{- 3} Q = 256 mC$

Since the net capacitance in upper arm and lower arm is same, so the charge gets divided equally between both the arms.

So, total charge flowing in the combination of lower arm of capacitors is-
$Q_{1} = \frac{Q}{2} Q_{1} = \frac{256 mC}{2} Q_{1} = 128 mC$

This is also the charge present on each of the lower two capacitors.

Similarly the charge on each of the upper two capacitors is also 128 mC

Now, voltage present on each of the lower capacitor is given by-
$\begin{array}{rcl} V & = & \frac{Q_{1}}{C} \\ = & \frac{128 mC}{2 mF} \\ = & 64 Volts \end{array}$

Similarly the voltage across each of the capacitors present in upper arm is also 64 Volts

(b)

When switch S is closed, it becomes a balanced condition of Wheatstone bridge and thus the potential difference between the two ends of the middle arm is zero.

It is now the similar case as of part (a). Hence charge on each of the capacitor is 128 mC and voltage on each of the capacitor is 64 Volts.

Step by step

Solved in 3 steps with 1 images

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.

Similar questions