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Application Of Kirchhoffs Rules Involving Capacitors

Practice to find the charges in all the capacitors by Kirchhoff's first and Second Law. Learn Kirchhoff’s Rules and Loop Circuits & Voltage Law with examples.

Kirchhoff's First Law of Capacitor

  • All charges that enter a given point in a circuit must leave that point because charge cannot build up at a point.

  • According to Kirchhoff's First Law, amount of charge entering a junction is same as amount of charge leaving the junction.

\(\sum Q_{\text{Enter}}=\sum Q_{\text{Leave}}\)

or  Q1 = Q2 + Q3

Illustration Questions

A total charge Q = 5 \(\mu C\) is entering at point P, as shown in figure. Determine the charge in C2 = 2 \(\mu F\) capacitor if charge in C1 = 3 \(\mu F\) capacitor is Q1 = 3 \(\mu C\).

A 8 \(\mu\)C

B 2 \(\mu\)C

C 4 \(\mu\)C

D 3 \(\mu\)C

×

According to Kirchhoff's First Law:

Amount of charge entering a junction is same as amount of charge leaving the junction.

image

\(\sum Q_{\text{Enter}}=\sum Q_{\text{Leave}}\)

 

image

Given: 

 Q = 5 \(\mu\)C,  Q1 = 3 \(\mu\)C

 Q = Q1 + Q2

\(\mu\)C = 3 \(\mu\)C + Q2 

or,  Q2 = 5 \(\mu\)C –  3 \(\mu\)C

or, Q2 = 2 \(\mu\)C

image

A total charge Q = 5 \(\mu C\) is entering at point P, as shown in figure. Determine the charge in C2 = 2 \(\mu F\) capacitor if charge in C1 = 3 \(\mu F\) capacitor is Q1 = 3 \(\mu C\).

image
A

\(\mu\)C

.

B

\(\mu\)C

C

\(\mu\)C

D

\(\mu\)C

Option B is Correct

Distribution of Charge by Kirchhoff's Law

  • Consider the circuit in which 3 capacitors having net charges Q1, Q2 and Q3 respectively are connected, as shown in figure.

According to Kirchhoff's First Law

  • Amount of charge entering a junction is same as amount of charge leaving the junction.
  • So, at point P

\(\sum Q_{\text{enter}}=\sum Q_{\text{leaving}}\)

Q3 = Q1 + Q2

Illustration Questions

Two capacitors C1 = 2 \(\mu\)F and C2 = 3 \(\mu\)F are connected with battery  \(\mathcal{E} =\)1 \(V\), as shown in figure. Calculate the  charge in   C2 = 3 \(\mu\)F capacitor when the charge through C1 is Q1= 2\(\mu\)C, given total charge Q = 5 \(\mu\)C.

A 6 \(\mu\)C

B 2 \(\mu\)C

C 3 \(\mu\)C

D 4 \(\mu\)C

×

According to Kirchhoff's First Law:

Amount of charge entering a junction is same as amount of charge leaving the junction.

image

At point P,

\(\sum Q_{\text{enter}}=\sum Q_{\text{leaving}}\)

or, Q = Q1 + Q2

or, \(\mu\)C = 2 \(\mu\)C + Q2 

or, Q2 = \(\mu\)C

image

Two capacitors C1 = 2 \(\mu\)F and C2 = 3 \(\mu\)F are connected with battery  \(\mathcal{E} =\)1 \(V\), as shown in figure. Calculate the  charge in   C2 = 3 \(\mu\)F capacitor when the charge through C1 is Q1= 2\(\mu\)C, given total charge Q = 5 \(\mu\)C.

image
A

\(\mu\)C

.

B

\(\mu\)C

C

\(\mu\)C

D

\(\mu\)C

Option C is Correct

Voltmeter

  • Voltmeter is a device used for measurement of voltage.
  • Voltmeter is always connected in parallel.

Illustration Questions

Choose the incorrect option. Given C1 = 1 \(\mu\)F,  C2 = 3 \(\mu\)F,  C3 = 3 \(\mu\)F and  C4 = 3 \(\mu\)F.

A V1  \(\neq\) V2

B V2  = V3

C V3  = V4

D V1  = 1 V

×

Charge stored by Capacitor is given as 

\(Q=CV\)

or, \(V=\dfrac {Q}{C}\)

image

Voltage across Capacitor C1

\(V_1=\dfrac {Q_1}{C_1}=\dfrac {1\mu C}{1\mu F}=1\; \text {V}\)

image

Voltage across Capacitor C2

\(V_2=\dfrac {Q_2}{C_2}=\dfrac {2\mu C}{3\mu F}=\dfrac {2}{3}\;\text{V}\)

 

image

Voltage across Capacitor C3

\(V_3=\dfrac {Q_3}{C_3}=\dfrac {1\;\mu C}{3\;\mu F}=\dfrac {1}{3}\;\text{V}\)

image

Voltage across Capacitor C4

\(V_4=\dfrac {Q_4}{C_4}=\dfrac {1\;\mu C}{3\;\mu F}=\dfrac {1}{3}\;\text{V}\)

image

Hence, \(V_1=1 V\)\(V_2=\dfrac {2}{3}V\)\(V_3=V_4=\dfrac {1}{3}V\)

So, option B is incorrect.

image

Choose the incorrect option. Given C1 = 1 \(\mu\)F,  C2 = 3 \(\mu\)F,  C3 = 3 \(\mu\)F and  C4 = 3 \(\mu\)F.

image
A

V1  \(\neq\) V2

.

B

V2  = V3

C

V3  = V4

D

V1  = 1 V

Option B is Correct

Kirchhoff's Second Law

  • Kirchhoff's second law states that the sum of potential differences across all elements around any closed circuit loop must be zero.

\(\displaystyle\sum_{\text{closed loop}}\Delta V = 0\)

  • This law is based upon principle of energy conservation because a charge that moves around any closed path and returns to its starting point, has zero potential energy [\(\Delta\)U = 0]

Case I : If capacitor is traversed from negative plate to positive plate.

\(\Delta V=+\dfrac {Q}{C}\)

Case II: If capacitor is traversed from positive plate to negative plate.

\(\Delta V=-\dfrac {Q}{C}\)

Illustration Questions

A single loop contains two capacitors and two sources of emf, as shown in figure. Choose the correct expression through Kirchhoff's second law.

A \(\mathcal{E}_1-\dfrac {Q}{C_2}-\mathcal{E}_2-\dfrac{Q}{C_1}=0\)

B \(-\mathcal{E}_1-\dfrac {Q}{C_2}+\mathcal{E}_2-\dfrac{Q}{C_2}=0\)

C \(\mathcal{E}_1-\dfrac {Q}{C_2}+\mathcal{E}_2+\dfrac{Q}{C_1}=0\)

D \(\mathcal{E}_1+\dfrac {Q}{C_2}-\mathcal{E}_2-\dfrac{Q}{C_1}=0\)

×

For Battery:

Moving from Positive to Negative terminal  \(\Delta V =-\mathcal{E}\)

Moving from Negative to Positive terminal \(\Delta V =+\mathcal{E}\)

image

For Capacitor:

Traversing from Negative to Positive plate \(\Delta V=+\dfrac {Q}{C}\)

Traversing from Positive to Negative plate \(\Delta V=-\dfrac {Q}{C}\)

image

By Kirchhoff's Second Law:

Traversing clockwise in the circuit

\(\mathcal{E}_1 -\dfrac{Q}{C_2}-\mathcal{E}_2-\dfrac{Q}{C_1}=0\)

image

A single loop contains two capacitors and two sources of emf, as shown in figure. Choose the correct expression through Kirchhoff's second law.

image
A

\(\mathcal{E}_1-\dfrac {Q}{C_2}-\mathcal{E}_2-\dfrac{Q}{C_1}=0\)

.

B

\(-\mathcal{E}_1-\dfrac {Q}{C_2}+\mathcal{E}_2-\dfrac{Q}{C_2}=0\)

C

\(\mathcal{E}_1-\dfrac {Q}{C_2}+\mathcal{E}_2+\dfrac{Q}{C_1}=0\)

D

\(\mathcal{E}_1+\dfrac {Q}{C_2}-\mathcal{E}_2-\dfrac{Q}{C_1}=0\)

Option A is Correct

Application of Kirchhoff's Second Law on a Circuit having Two Loops

  • Kirchhoff's second law states that the sum of potential differences across all elements around any closed circuit loop must be zero.

\(\displaystyle\sum_{\text{closed loop}}\Delta V = 0\)

  • This law is based upon principle of energy conservation because a charge that moves around any closed path and returns to its starting point, has zero potential energy [\(\Delta\)U = 0].

Case I : If capacitor is traversed from negative plate to positive plate.

\(\Delta V=+\dfrac {Q}{C}\)

Case II: If capacitor is traversed from positive plate to negative plate.

\(\Delta V=-\dfrac {Q}{C}\)

  • Consider the circuit, as shown in figure.

  • Using Kirchhoff's Second Law

For Loop 1 :\(\mathcal{E}_1-\dfrac {Q_1}{C_2}-\mathcal{E}_2+\dfrac {Q}{C_1}=0\)

For Loop 2 : \(-\dfrac {Q_2}{C_3}+\dfrac {Q_1}{C_2}=0\)

Illustration Questions

Which option is correct according to Kirchhoff's Second Law?

A For Loop 1 :\(\mathcal{E}_1-\dfrac {Q_1}{C_1}=0\) For Loop 2 : \(-\dfrac {Q_2}{C_2}+\dfrac {Q_1}{C_1}=0\)

B For Loop 1 :\(\mathcal{E}_1+\dfrac {Q_1}{C_1}=0\) For Loop 2 : \(\dfrac {Q_2}{C_2}+\dfrac {Q_1}{C_1}=0\)

C For Loop 1 :\(\mathcal{E}_1-\dfrac {Q_1}{C_1}=0\) For Loop 2 : \(-\dfrac {Q_2}{C_2}-\dfrac {Q_1}{C_1}=0\)

D For Loop 1 :\(\mathcal{E}_1+\dfrac {Q_1}{C_1}=0\) For Loop 2 : \(-\dfrac {Q_2}{C_2}-\dfrac {Q_1}{C_1}=0\)

×

For Battery:

Moving from Positive to Negative terminal  \(\Delta V =-\mathcal{E}\)

Moving from Negative to Positive terminal \(\Delta V =+\mathcal{E}\)

image

For Capacitor:

Traversing from Negative to Positive plate \(\Delta V=+\dfrac {Q}{C}\)

Traversing from Positive to Negative plate \(\Delta V=-\dfrac {Q}{C}\)

image

By using Kirchhoff's Second Law:

Traversing in clockwise direction

For Loop 1 :\(\mathcal{E}_1-\dfrac {Q_1}{C_1}=0\)

For Loop 2 : \(-\dfrac {Q_2}{C_2}+\dfrac {Q_1}{C_1}=0\)

image

Hence, Option (A) is correct.

image

Which option is correct according to Kirchhoff's Second Law?

image
A

For Loop 1 :\(\mathcal{E}_1-\dfrac {Q_1}{C_1}=0\)

For Loop 2 : \(-\dfrac {Q_2}{C_2}+\dfrac {Q_1}{C_1}=0\)

.

B

For Loop 1 :\(\mathcal{E}_1+\dfrac {Q_1}{C_1}=0\)

For Loop 2 : \(\dfrac {Q_2}{C_2}+\dfrac {Q_1}{C_1}=0\)

C

For Loop 1 :\(\mathcal{E}_1-\dfrac {Q_1}{C_1}=0\)

For Loop 2 : \(-\dfrac {Q_2}{C_2}-\dfrac {Q_1}{C_1}=0\)

D

For Loop 1 :\(\mathcal{E}_1+\dfrac {Q_1}{C_1}=0\)

For Loop 2 : \(-\dfrac {Q_2}{C_2}-\dfrac {Q_1}{C_1}=0\)

Option A is Correct

Illustration Questions

Choose the correct option according to Kirchhoff's Law.

A For Loop 1 : \(\mathcal{E}_1-\dfrac {Q_1}{C_1}=0\) For Loop 2 : \(-\mathcal{E}_2+\dfrac {Q_2}{C_2}=0\)

B For Loop 1 : \(-\mathcal{E}_1-\dfrac {Q_1}{C_1}=0\) For Loop 2 : \(\mathcal{E}_2+\dfrac {Q_2}{C_2}=0\)

C For Loop 1 : \(-\mathcal{E}_1-\dfrac {Q_1}{C_1}=0\) For Loop 2 : \(-\mathcal{E}_2+\dfrac {Q_2}{C_2}=0\)

D For Loop 1 : \(\mathcal{E}_1-\dfrac {Q_1}{C_1}=0\) For Loop 2 : \(-\mathcal{E}_2-\dfrac {Q_2}{C_2}=0\)

×

For Battery:

Moving from Positive to Negative terminal  \(\Delta V =-\mathcal{E}\)

Moving from Negative to Positive terminal \(\Delta V =+\mathcal{E}\)

For Capacitor:

Traversing from Negative to Positive plate \(\Delta V=+\dfrac {Q}{C}\)

Traversing from Positive to Negative plate \(\Delta V=-\dfrac {Q}{C}\)

By using Kirchhoff's Law:

For Loop 1 : \(\mathcal{E}_1-\dfrac {Q_1}{C_1}=0\)

image

By using Kirchhoff's Law:

For Loop 2 : \(-\mathcal{E}_2+\dfrac {Q_2}{C_2}=0\)

image

Hence, option (A) is correct.

Choose the correct option according to Kirchhoff's Law.

image
A

For Loop 1 : \(\mathcal{E}_1-\dfrac {Q_1}{C_1}=0\)

For Loop 2 : \(-\mathcal{E}_2+\dfrac {Q_2}{C_2}=0\)

.

B

For Loop 1 : \(-\mathcal{E}_1-\dfrac {Q_1}{C_1}=0\)

For Loop 2 : \(\mathcal{E}_2+\dfrac {Q_2}{C_2}=0\)

C

For Loop 1 : \(-\mathcal{E}_1-\dfrac {Q_1}{C_1}=0\)

For Loop 2 : \(-\mathcal{E}_2+\dfrac {Q_2}{C_2}=0\)

D

For Loop 1 : \(\mathcal{E}_1-\dfrac {Q_1}{C_1}=0\)

For Loop 2 : \(-\mathcal{E}_2-\dfrac {Q_2}{C_2}=0\)

Option A is Correct

Illustration Questions

Calculate potential difference across point a and b.

A 5 V

B 16 V

C 8 V

D 32 V

×

For Battery

Moving from Positive to Negative terminal  \(\Delta V =-\mathcal{E}\)

Moving from Negative to Positive terminal \(\Delta V =+\mathcal{E}\)

image

For Capacitor

Traversing from Negative to Positive plate \(\Delta V=+\dfrac {Q}{C}\)

Traversing from Positive to Negative plate \(\Delta V=-\dfrac {Q}{C}\)

image

Using Kirchhoff's Second Law

For Loop 1

Traversing in clockwise direction

\(20V-\dfrac {Q}{4\times10^{-6}}-\dfrac {Q_2}{2\times10^{-6}}=0\)

or, \(80\times10^{-6}-Q-2Q_2=0\) ...(i)

image

Using Kirchhoff's Second Law

For Loop 2

Traversing in clockwise direction

\(\dfrac {-Q_1}{4\times10^{-6}}+\dfrac {Q_2}{2\times10^{-6}}=0\)

\(\dfrac {Q_1}{4\times10^{-6}}=\dfrac {Q_2}{2\times10^{-6}}\)

or, \(Q_1=2Q_2\) ...(ii)

image

Using Kirchhoff's First Law

At Junction J

\(Q=Q_1+Q_2\)  .....(iii)

image

Solving (i) and (iii)

\(80\times10^{-6}-(Q_1+Q_2)-2Q_2=0\)

or, \(80\times10^{-6}-Q_1-3Q_2=0\)  ...(iv)

image

Solving (ii) and (iv)

\(80\times10^{-6}-(2Q_2)-(3Q_2)=0\)     \(\bigg[Q_1=2Q_2\bigg]\)

or,   \(80\times10^{-6}-5Q_2=0\)

 \(Q_2=\dfrac {80×10^{-6}}{5}=16\times10^{-6}C\)

image

Potential across 2 \(\mu\)F capacitor

\(V_{2 \;\mu F}=\dfrac {Q_2}{2\;\mu F}\)

or, \(V_{2 \;\mu F}=\dfrac {16\times10^{-6}}{2\times 10^{-6}}=8V\)

image

Hence, potential across a and b

\(V_{ab}=V_{2\;\mu F}=8V\)

image

Hence, option (C) is correct.

image

Calculate potential difference across point a and b.

image
A

5 V

.

B

16 V

C

8 V

D

32 V

Option C is Correct

Short Circuit

  • When two points of a circuit are connected together by a conducting wire, they are said to be short circuited.
  • The connecting wire is assumed to have zero resistance.

  • Consider a circuit with 3 capacitors and a battery connected, as shown in figure.
  • If capacitor (C3) is short circuited, then potential difference across capacitor drops to zero.

\(\Delta V_{C_3}=0\)

  • Since C2 and C3 are connected in parallel, so potential across both, are same.

\(\Delta V_{C_3}=\Delta V_{C_2}\)

Hence, \(\Delta V_{C_2}=0\)

  • Hence, no charge will flow in C2 and C3.
  • By Kirchhoff's Law for shown circuit.

\(\mathcal{E}_1- \dfrac {Q_1}{C_1}=0\)

or, \(\dfrac {Q_1}{C_1}=\mathcal{E}_1\)

Illustration Questions

Find the charges in all the capacitors C1 = 1 \(\mu\)F, C2 = 2 \(\mu\)F  and C3 = 3 \(\mu\)F, connected with a battery \(\mathcal{E}_1=7V\), when C3 is short circuited.

A 0, 7\(\mu\)C, 7\(\mu\)C

B 0, 0, 7\(\mu\)C

C 7\(\mu\)C, 0, 0

D 0, 7\(\mu\)C, 0

×

Since, C3 is short circuited so, voltage across C3 is zero.

\(\Delta\;V_{C_3}=0\)

Hence, no charge flows in C3 .

\(Q_3=0\)

image image

Since, C2 and C3 are connected in parallel and  potential across parallel is same.

So, \(\Delta \;V_{C_3}=\Delta \;V_{C_2}=0\ Volt\) 

Also, no charge flows in capacitor C.

\(Q_2=0\)

image

The circuit reduces to, as shown in figure. 

By Kirchhoff's Second Law

\(\mathcal{E}_1-\dfrac {Q_1}{C_1}=0\)

\(\mathcal{E}_1=\dfrac {Q_1}{C_1}\)

\(Q_1=\mathcal{E}_1C_1\)

image image

Charge in Capacitor C1

\(Q_1=\mathcal{E}_1C_1\)

or, \(Q_1=7\times1\mu F\)

or, \(Q_1=7\mu C\)

image

Find the charges in all the capacitors C1 = 1 \(\mu\)F, C2 = 2 \(\mu\)F  and C3 = 3 \(\mu\)F, connected with a battery \(\mathcal{E}_1=7V\), when C3 is short circuited.

image
A

0, 7\(\mu\)C, 7\(\mu\)C

.

B

0, 0, 7\(\mu\)C

C

7\(\mu\)C, 0, 0

D

0, 7\(\mu\)C, 0

Option C is Correct

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