Informative line

### 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

#### 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.

$$\sum Q_{\text{Enter}}=\sum Q_{\text{Leave}}$$

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

### 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

$$\mu$$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

#### 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.

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

### 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

$$\mu$$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.

#### 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, $$V=\dfrac {Q}{C}$$

Voltage across Capacitor C1

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

Voltage across Capacitor C2

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

Voltage across Capacitor C3

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

Voltage across Capacitor C4

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

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

So, option B is incorrect.

### 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

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.

$$\Delta V=-\dfrac {Q}{C}$$

#### 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 $$\Delta V =+\mathcal{E}$$

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 Kirchhoff's Second Law:

Traversing clockwise in the circuit

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

### 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

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

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 : $$-\dfrac {Q_2}{C_3}+\dfrac {Q_1}{C_2}=0$$

#### 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 $$\Delta V =+\mathcal{E}$$

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 Second Law:

Traversing in clockwise direction

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

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

Hence, Option (A) is correct.

### 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 : $$-\dfrac {Q_2}{C_2}-\dfrac {Q_1}{C_1}=0$$

Option A is Correct

#### 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 : $$\mathcal{E}_1-\dfrac {Q_1}{C_1}=0$$

By using Kirchhoff's Law:

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

Hence, option (A) is correct.

### 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 : $$-\mathcal{E}_2-\dfrac {Q_2}{C_2}=0$$

Option A is Correct

#### 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 $$\Delta V =+\mathcal{E}$$

For Capacitor

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

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

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)

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)

Using Kirchhoff's First Law

At Junction J

$$Q=Q_1+Q_2$$  .....(iii)

Solving (i) and (iii)

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

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

Solving (ii) and (iv)

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

or,   $$80\times10^{-6}-5Q_2=0$$

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

Potential across 2 $$\mu$$F capacitor

$$V_{2 \;\mu F}=\dfrac {Q_2}{2\;\mu F}$$

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

Hence, potential across a and b

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

Hence, option (C) is correct.

### Calculate potential difference across point a and b.

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, $$\dfrac {Q_1}{C_1}=\mathcal{E}_1$$

#### 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 .

$$Q_3=0$$

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.

$$Q_2=0$$

The circuit reduces to, as shown in figure.

By Kirchhoff's Second Law

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

$$\mathcal{E}_1=\dfrac {Q_1}{C_1}$$

$$Q_1=\mathcal{E}_1C_1$$

Charge in Capacitor C1

$$Q_1=\mathcal{E}_1C_1$$

or, $$Q_1=7\times1\mu F$$

or, $$Q_1=7\mu C$$

### 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

Option C is Correct