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Physics Chapter 11 : Topic 13
Kirchhoff’s Laws in Circuits
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Kirchhoff’s laws are mainly used to analyze circuits that:
Reasoning: Kirchhoff’s laws are especially useful for complex circuits containing multiple loops and branches.
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Kirchhoff’s first law is applied at:
Reasoning: Kirchhoff’s first law deals with current entering and leaving a junction.
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According to Kirchhoff’s first law, the algebraic sum of currents at a junction is:
Reasoning: Current entering a junction equals current leaving it due to conservation of charge.
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A circuit has currents of $\mathbf{3\ A}$ and $\mathbf{2\ A}$ entering a junction. The current leaving the junction is:
Reasoning: According to Kirchhoff’s first law, total incoming current equals total outgoing current.
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Kirchhoff’s second law is applied to:
Reasoning: Kirchhoff’s second law states that the total voltage change around a closed loop is zero.
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When solving a circuit using Kirchhoff’s laws, the first step is usually to:
Reasoning: Assuming current directions allows equations to be written. A negative result indicates the actual direction is opposite.
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A negative value of current obtained from Kirchhoff’s equations means:
Reasoning: The sign only indicates that the initial assumed current direction was incorrect.
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In applying Kirchhoff’s second law, the voltage rise across a battery is considered:
Reasoning: Moving through a battery from negative to positive terminal increases potential energy.
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A resistor crossed in the direction of current gives a potential change that is:
Reasoning: Moving through a resistor in the direction of current causes a voltage drop.
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A $\mathbf{12\ V}$ battery is connected to a $\mathbf{4\ \Omega}$ resistor. The current in the circuit is:
Reasoning: Using Ohm’s law $\mathbf{\Longrightarrow}$ $\mathbf{I =}$ $\frac{\mathbf{V}}{\mathbf{R}}\mathbf{=}\frac{\mathbf{12}}{\mathbf{4}}\mathbf{= 3\ A}$
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A loop contains a $\mathbf{10\ V}$ battery and two resistors of $\mathbf{2\ \Omega}$ and $\mathbf{3\ \Omega}$ in series. The current is:
Reasoning: Total resistance $\mathbf{= R = 2 + 3 = 5\ \Omega}$ $\mathbf{\Longrightarrow}$ $\mathbf{I =}$ $\frac{\mathbf{V}}{\mathbf{R}}\mathbf{=}\frac{\mathbf{10}}{\mathbf{5}}\mathbf{= 2\ A}$
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In a multi-loop circuit, Kirchhoff’s laws provide:
Reasoning: Each junction and loop provides equations needed to solve unknown currents.
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Two batteries of $\mathbf{6\ V}$ and $\mathbf{4\ V}$ oppose each other in a circuit with resistance $\mathbf{5\ \Omega}$. The current is:
Reasoning: Net $\mathbf{e.m.f = 6 - 4 = 2\ V}$ $\mathbf{\Longrightarrow}$ $\mathbf{I =}$ $\frac{\mathbf{V}}{\mathbf{R}}\mathbf{=}\frac{\mathbf{2}}{\mathbf{5}}\mathbf{= 0.4\ A}$
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When two batteries aid each other in a circuit, the total $\mathbf{emf}$ is:
Reasoning: Batteries connected in the same direction add their potential rises.
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When two batteries oppose each other, the effective $\mathbf{e.m.f.}$ is:
Reasoning: Opposing sources reduce the net energy supplied to the circuit.
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A circuit has a battery of $\mathbf{20\ V}$ and resistors of $\mathbf{4\ \Omega}$ and $\mathbf{6\ \Omega}$ in series. The voltage drop across $\mathbf{4\ \Omega}$ resistor is:
Reasoning: Total resistance $\mathbf{= 10\ \Omega}$Current $\mathbf{= I = V\ /\ R}\mathbf{= 20\ /\ 10 = 2\ A}$ Voltage drop $\mathbf{= V}\mathbf{= IR = 2 \times 4 = 8\ V}$
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Kirchhoff’s second law for a simple loop containing a battery and resistor gives:
Reasoning: The energy supplied by the battery equals the energy lost in the resistor.
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A current entering a junction is represented as positive and leaving current as negative. Kirchhoff’s first law becomes:
Reasoning: The algebraic sum of currents at a junction must equal zero.
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The main reason for using Kirchhoff’s laws instead of Ohm’s law alone is:
Reasoning: Kirchhoff’s laws allow simultaneous equations to solve complicated networks.
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A circuit loop has a total $\mathbf{emf}$ of $\mathbf{15\ V}$ and resistor drops of $\mathbf{5\ V\ }$and $\mathbf{10\ V}$. This satisfies:
Reasoning: Voltage rise equals total voltage drop. $\mathbf{\Longrightarrow}$ $\mathbf{15 - 5 - 10 = 0}$
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If a resistor has current $\mathbf{2\ A}$ and resistance $\mathbf{5\ \Omega}$, its voltage drop is:
Reasoning: Using Ohm’s law $\mathbf{\Longrightarrow}$ $\mathbf{V = IR = 2(5) = 10\ V}$
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In a circuit with multiple loops, the number of loop equations required depends on:
Reasoning: Each independent loop provides a Kirchhoff voltage equation.
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A resistor network cannot be solved directly by series and parallel formulas when:
Reasoning: Bridge circuits often require Kirchhoff’s laws because resistors are not purely series or parallel.
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A current of $\mathbf{4\ A}$ enters a junction and divides into currents of $\mathbf{1\ A}$ and $\mathbf{2\ A}$. The third branch current is:
Reasoning: Total outgoing current must equal incoming current ; $\mathbf{4 = 1 + 2 +}\mathbf{I}$
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Kirchhoff’s laws are valid because of:
Reasoning: First law follows from charge conservation; second law follows from energy conservation.
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A loop contains a battery of $\mathbf{18\ V}$ and resistors of $\mathbf{3\ \Omega}$ and $\mathbf{6\ \Omega}$. The current is:
Reasoning: Total resistance $\mathbf{= 9\ \Omega}$ $\mathbf{\Longrightarrow}$ $\mathbf{I =}$ $\frac{\mathbf{18}}{\mathbf{9}}\mathbf{= 2\ A}$
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While applying Kirchhoff’s second law, changing the assumed loop direction:
Reasoning: Direction choice only affects sign convention.
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A circuit equation obtained from Kirchhoff’s second law represents conservation of:
Reasoning: The total energy gained and lost around a loop must balance.
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If the calculated current in a branch is zero, it means:
Reasoning: A zero current solution indicates no charge flow through that branch under given conditions.
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The final purpose of using Kirchhoff’s laws in circuit analysis is to determine:
Reasoning: Kirchhoff’s laws provide equations to find unknown currents and voltages in circuits.
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