Q: The potential difference \$(V_A-V_B)\$ between the points A and B in the given figure is.

Given : \$I = 2\$ A Using Kirchhoff's law \$V_A - I(2) - 3 - I(1) - V_B = 0\$\$\therefore\$ \$V_A - 2\times 2 - 3 - 2\times 1 -V_B = 0\$ OR \$V_A - 9 - V_B = 0\$ \$\implies V_A - V_B = +9\$ V

Q: Draw a circuit diagram of an electric circuit containing a cell, a key, an ammeter, a resistor of \$4\Omega\$ in series with a combination of two resistors ( \$8\Omega\$ each ) in parallel and a voltmeter across parallel combination. Each of them dissipate maximum energy and canwithstand a maximum power of \$16W\$ without melting. Find the maximum current that can flow through the three resistors.

Maximum current through \$4\Omega\$ resister \$=\sqrt{\dfrac{P}{R}}\$ \$=\sqrt{\dfrac{16}{4}}=2\ A\$MAximum current through each \$8\Omega\$ resister \$=\dfrac{1}{2}\times2=1\ A\$

Q: The figure below shows currents in a part of electric circuit. The current \$ i \$ is

According to Kirchhoff's first lawAt junction \$A, i_{AB}=2+2=4\ A\$At junction \$B, i_{AB}=i_{BC}-1=3\ A\$At junction \$C, i=i_{BC}-1.3=3-1.3=1.7\ amp\$

Q: Kirchoffs 1st and 2nd laws are based on conservation of

Kirchhoff's \$1\$st law states that the total amount of current coming at a junction must be equal to the total amount of current going away from it i.e. the total charge must be conserved. Thus Kirchhoff's \$1\$st law is based on the conservation of charge.Kirchhoff's \$2\$nd law states that the sum of the potential drop across all the components in a loop must be zero. Thus Kirchhoff's \$2\$nd law is based on the conservation of energy.

Q: In the circuit shown, the value of currents \$I_1, I_2\$ and \$I_3\$ are.

Applying Kirchoff's voltage law, In loop. \$I\$ \$ = -27 -6I_2 - 2I_1 + 24 = 0\$\$\Rightarrow 6I_2 + 2I_1 = -3\$ ...(1)In loop- \$II\$\$-27 - 6I_2 + 4I_3 = 0\$\$6I_2 - 4I_3 = -27\$ ...(2)Junction '\$p\$'\$I_1 = I_2 + I_3\$ ...(3)Solving equation (1), (2) and (3) we get,\$I_1 = 3A, \,I_2 = -3/2 A, \,I_3 = 9/2A\$

Question 1

The potential difference $(V_A-V_B)$ between the points A and B in the given figure is.

Accepted Answer

Given :   $I = 2$ A           Using Kirchhoff's law       $V_A  - I(2) - 3 - I(1) - V_B = 0$$\therefore$    $V_A - 2\times 2 - 3  - 2\times 1 -V_B = 0$             OR     $V_A - 9 - V_B = 0$               $\implies V_A - V_B = +9$ V

Question 2

Draw a circuit diagram of an electric circuit containing a cell, a key, an ammeter, a resistor of $4\Omega$ in series with a combination of two resistors ( $8\Omega$ each ) in parallel and a voltmeter across parallel combination. Each of them dissipate maximum energy and canwithstand a maximum power of $16W$ without melting. Find the maximum current that can flow through the three resistors.

Accepted Answer

Maximum current through $4\Omega$ resister $=\sqrt{\dfrac{P}{R}}$                                                                $=\sqrt{\dfrac{16}{4}}=2\ A$MAximum current through each $8\Omega$ resister $=\dfrac{1}{2}\times2=1\ A$

Question 3

The figure below shows currents in a part of electric circuit. The current $ i $ is

Accepted Answer

According to Kirchhoff's first lawAt junction $A, i_{AB}=2+2=4\ A$At junction $B, i_{AB}=i_{BC}-1=3\ A$At junction $C, i=i_{BC}-1.3=3-1.3=1.7\ amp$

Question 4

Kirchoffs 1st and 2nd laws are based on conservation of

Accepted Answer

Kirchhoff's $1$st law states that the total amount of current coming at a junction must be equal to the total amount of current going away from it i.e. the total charge must be conserved. Thus Kirchhoff's $1$st law is based on the conservation of charge.Kirchhoff's $2$nd law states that the sum of the potential drop across all the components in a loop must be zero. Thus Kirchhoff's $2$nd law is based on the conservation of energy.

Question 5

State and explain Kirchhoff's Laws in the current electricity.

Accepted Answer

Kirchhoff's laws: (i) The algebraic sum of current meeting at an junction in a circuit is zero. In this law, the current flowing towards the junction are considered as $+$ve and those flowing away from the junction as $-$ve currents.As shown in the figure, we have     ${i}_{1} - {i}_{2} - {i}_{3} + {i}_{4} - {i}_{5} = 0$or  ${i}_{1} + {i}_{4} = {i}_{2} + {i}_{3} + {i}_{5}$(ii) In any closed mesh (or loop) of an electrical circuit, be algebraic sum of the product of the currents and resistances is equal to the total e.m.f. of the mesh.     If we go along the direction of conventional current, be potential difference will be taken as negative and opposite to it will be positive.    Inside the cell, if we move from low to high potential along the direction of conventional current the e.m.f. will be positive.For loop 1,          ${E}_{2} - {i}_{2}{R}_{2} - \left({i}_{1} + {i}_{2}\right) {R}_{3} = 0$or       ${E}_{2} = {i}_{2}{R}_{2} + \left({i}_{1} + {i}_{2}\right) {R}_{3}$For loop 2,         ${i}_{2}{R}_{2} - {E}_{2} - {i}_{1}{R}_{1} + {E}_{1} = 0$or      ${E}_{1} - {E}_{2} = {i}_{1}{R}_{1} - {i}_{2}{R}_{2}$

Question 6

A switch S is closed in the circuit shown in figure, current passed through point A is?

Accepted Answer

Toppr Ask Answer

Question 7

In the circuit shown, the value of currents $I_1, I_2$ and $I_3$ are.

Accepted Answer

Applying Kirchoff's voltage law, In loop. $I$ $ = -27 -6I_2 - 2I_1 + 24 = 0$$\Rightarrow 6I_2 + 2I_1 = -3$     ...(1)In loop- $II$$-27 - 6I_2 + 4I_3 = 0$$6I_2 - 4I_3 = -27$      ...(2)Junction '$p$'$I_1 = I_2 + I_3$      ...(3)Solving equation (1), (2) and (3) we get,$I_1 = 3A, \,I_2 = -3/2 A, \,I_3 = 9/2A$

Question 8

Consider the circuit shown in the figure. The current $I_3$ is equal to

Accepted Answer

Suppose current through different paths of the circuit is as follows.After applying $KVL$, for loop (1) and (2)We get $28i_1=-6-8\Rightarrow i_1=-\dfrac 12\ A$and $54i_2=-6-12\Rightarrow i_2=-\dfrac 13\ A$Hence $i_3=i_1+i_2=-\dfrac 56\ A$

Question 9

The figure shows a network of currents. The magnitude of currents is shown here. The current $i$ will be

Accepted Answer

Kirchhoff's junction rule states that at any junction ( node ) in an electrical circuit, the sum of the currents flowing into that junction is equal to the sum of the currents flowing out of that junction.$i_{in}=i_{out}$so $i_{in}=15+3=18\,A$and $i_{out}=i+3$so $18=i+3\implies i=15\,A$

Question 10

State Kirchoff's current law and also give the sign convention for current.

Accepted Answer

Kirchhoff's current law states that at any node(junction) in an electric circuit, the sum currents flowing in to that node is equal to the sum of current flowing out of that node.Sign convention:- current entering at any junction is considered as positive.                        and current leaving the junction is considered as negative.In given circuit,By Kirchhoff's current law,$i_2+i_3=i_1+i_4$

Revise with Concepts

Kirchoff's Law

Learn with Videos

Quick Summary With Stories

Kirchhoff's Voltage Law

Conservation of Energy by Kirchoff's Law

Problems based on Kirchhoff's law

The potential difference $(V_{A} - V_{B})$ between the points A and B in the given figure is.

The figure below shows currents in a part of electric circuit. The current $i$ is

Kirchoffs 1st and 2nd laws are based on conservation of

State and explain Kirchhoff's Laws in the current electricity.

A switch S is closed in the circuit shown in figure, current passed through point A is?

In the circuit shown, the value of currents $I_{1}, I_{2}$ and $I_{3}$ are.

Consider the circuit shown in the figure. The current $I_{3}$ is equal to

The figure shows a network of currents. The magnitude of currents is shown here. The current $i$ will be

State Kirchoff's current law and also give the sign convention for current.

Revise with Concepts

Kirchoff's Law

Learn with Videos

Quick Summary With Stories

Kirchhoff's Voltage Law

Conservation of Energy by Kirchoff's Law

Problems based on Kirchhoff's law

The potential difference (VA​−VB​) between the points A and B in the given figure is.

The figure below shows currents in a part of electric circuit. The current i is

Kirchoffs 1st and 2nd laws are based on conservation of

State and explain Kirchhoff's Laws in the current electricity.

A switch S is closed in the circuit shown in figure, current passed through point A is?

In the circuit shown, the value of currents I1​,I2​ and I3​ are.

Consider the circuit shown in the figure. The current I3​ is equal to

The figure shows a network of currents. The magnitude of currents is shown here. The current i will be

State Kirchoff's current law and also give the sign convention for current.

The potential difference $(V_{A} - V_{B})$ between the points A and B in the given figure is.

The figure below shows currents in a part of electric circuit. The current $i$ is

In the circuit shown, the value of currents $I_{1}, I_{2}$ and $I_{3}$ are.

Consider the circuit shown in the figure. The current $I_{3}$ is equal to

The figure shows a network of currents. The magnitude of currents is shown here. The current $i$ will be