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.

- 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 Q_{1} = Q_{2} + Q_{3}

A 8 \(\mu\)C

B 2 \(\mu\)C

C 4 \(\mu\)C

D 3 \(\mu\)C

- Consider the circuit in which 3 capacitors having net charges Q
_{1}, Q_{2}and Q_{3}respectively are connected, as shown in figure.

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

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

Q_{3} = Q_{1} + Q_{2}

A 6 \(\mu\)C

B 2 \(\mu\)C

C 3 \(\mu\)C

D 4 \(\mu\)C

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

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

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

A V1 \(\neq\) V2

B V2 = V3

C V3 = V4

D V1 = 1 V

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

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

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

- 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 (C
_{3}) is short circuited, then potential difference across capacitor drops to zero.

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

- Since C
_{2}and C_{3}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 C
_{2}and C_{3}. - By Kirchhoff's Law for shown circuit.

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

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

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