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Physics Chapter 11 : Topic 6
Electromotive Force (e.m.f.)
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The electromotive force $\mathbf{e.m.f.}$ of a source is defined as:
Reasoning: $\mathbf{e.m.f.}$ is the energy transferred by a source per unit charge to move charge around a complete circuit.
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The mathematical expression for $\mathbf{e.m.f.}$ is:
Reasoning: $\mathbf{e.m.f.}$ is energy transferred $\mathbf{W}$ per unit charge $\mathbf{Q}$.
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The SI unit of electromotive force is:
Reasoning: One volt equals one joule per coulomb $\mathbf{(1\ V = 1\ J\ /\ C)}$.
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A battery has an $\mathbf{e.m.f.}$ of $\mathbf{12\ V}$. This means that the battery:
Reasoning: $\mathbf{e.m.f.}$ represents energy supplied per coulomb of charge.
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A source transfers $\mathbf{100\ J}$ of energy to move $\mathbf{20\ C}$ of charge around a circuit. Its $\mathbf{e.m.f.}$ is:
Reasoning: $\mathbf{E =}$ $\frac{\mathbf{W}}{\mathbf{Q}}\mathbf{=}\frac{\mathbf{100}}{\mathbf{20}}\mathbf{= 5\ V}$
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The $\mathbf{e.m.f.}$ of a source is measured when:
Reasoning: $\mathbf{e.m.f.}$ is the energy supplied per charge in driving charge around a complete circuit.
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Which device converts chemical energy into electrical energy?
Reasoning: A battery uses chemical reactions to provide energy to charges.
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A $\mathbf{1.5\ V}$ cell transfers $\mathbf{3\ J}$ of energy. The charge moved is:
Reasoning: $\mathbf{E =}$ $\frac{\mathbf{W}}{\mathbf{Q}}$ $\mathbf{\Longrightarrow}$ So, $\mathbf{Q =}$ $\frac{\mathbf{W}}{\mathbf{E}}\mathbf{=}\frac{\mathbf{3}}{\mathbf{1.5}}\mathbf{= 2C}$
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The main difference between $\mathbf{e.m.f.}$ and potential difference is that:
Reasoning: $\mathbf{e.m.f.}$ represents energy provided by the source; potential difference represents energy transferred elsewhere.
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A source with greater $\mathbf{e.m.f.}$ can:
Reasoning: Larger $\mathbf{e.m.f.}$ means more energy supplied to each coulomb.
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A battery has an $\mathbf{e.m.f.\ }$of $\mathbf{9\ V}$. The energy supplied to $\mathbf{4\ C}$ of charge is:
Reasoning: $\mathbf{W = EQ = (9)(4) = 36\ J}$
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A cell does not actually produce charge. Instead, it:
Reasoning: Charges already exist in conductors; the cell provides energy to drive them.
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The symbol commonly used for electromotive force is:
Reasoning: $\mathbf{e.m.f.}$ is represented by $\mathbf{E}$ or the Greek letter epsilon $\mathbf{(\varepsilon)}$.
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If the energy supplied by a source doubles while charge remains constant, its $\mathbf{e.m.f.}$ will:
Reasoning: $\mathbf{E =}$ $\frac{\mathbf{W}}{\mathbf{Q}}$ $\mathbf{\Longrightarrow}$ If $\mathbf{W}$ doubles and $\mathbf{Q}$ remains constant, $\mathbf{E}$ doubles.
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If the charge moved through a source doubles while energy supplied remains constant, the $\mathbf{e.m.f.}$ becomes:
Reasoning: $\mathbf{E =}$ $\frac{\mathbf{W}}{\mathbf{Q}}$ $\mathbf{\Longrightarrow}$ Increasing charge reduces energy per charge.
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A source with $\mathbf{e.m.f.\ \ 2\ V}$ transfers energy to charge in such a way that:
Reasoning: $\mathbf{1\ V = 1\ J\ /\ C}$ $\mathbf{\Longrightarrow}$ So, $\mathbf{2\ V}$ means $\mathbf{2\ J}$ per coulomb.
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Which statement about $\mathbf{e.m.f.}$ is correct?
Reasoning: $\mathbf{e.m.f.}$ describes energy transfer per unit charge.
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A battery supplies $\mathbf{240\ J}$ of energy while moving $\mathbf{30\ C}$ of charge. Its $\mathbf{e.m.f.}$ is:
Reasoning: $\mathbf{E =}$ $\frac{\mathbf{W}}{\mathbf{Q}}\mathbf{=}\frac{\mathbf{240}}{\mathbf{30}}\mathbf{= 8\ V}$
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The work done by a source in moving charge around a complete circuit is related to:
Reasoning: Energy supplied $\mathbf{(W = EQ)}$.
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A battery rated at $\mathbf{6\ V}$ compared with a battery rated at $\mathbf{3\ V}$ provides:
Reasoning: A $\mathbf{6\ V}$ battery provides $\mathbf{6\ J\ /\ C\ }$, while a $\mathbf{3\ V}$ battery provides $\mathbf{3\ J\ /\ C}$.
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The $\mathbf{e.m.f.}$ of a cell is independent of:
Reasoning: $\mathbf{e.m.f.}$ is a property of the source and does not depend on the amount of charge moved.
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When a battery is connected to a circuit, its role is to:
Reasoning: The source supplies energy that drives charges through the circuit.
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A source transfers $\mathbf{50\ J}$ energy to $\mathbf{10\ C}$ charge. The potential generated by the source is:
Reasoning: $\mathbf{E =}$ $\frac{\mathbf{W}}{\mathbf{Q}}\mathbf{=}\frac{\mathbf{50}}{\mathbf{10}}\mathbf{= 5\ V}$
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A voltmeter connected across an ideal cell measures its:
Reasoning: An ideal voltmeter measures the potential difference, which equals $\mathbf{e.m.f.}$ for an ideal cell.
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The term “electromotive force” is misleading because:
Reasoning: $\mathbf{e.m.f.}$ is energy per unit charge, not a physical force.
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If a source moves $\mathbf{50\ C}$ charge and supplies $\mathbf{150\ J}$ energy, its $\mathbf{e.m.f.}$ is:
Reasoning: $\mathbf{E =}$ $\frac{\mathbf{150}}{\mathbf{50}}\mathbf{= 3\ }\mathbf{V}$
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Which situation represents a source with high $\mathbf{e.m.f.}$?
Reasoning: High $\mathbf{e.m.f.}$ means greater energy transfer per unit charge.
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The energy transferred by a source depends on:
Reasoning: $\mathbf{(W = EQ)}$
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A $\mathbf{12\ V}$ battery moves $\mathbf{5\ C}$ of charge. The energy supplied is:
Reasoning: $\mathbf{W = EQ = (12)(5) = 60\ J}$
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The correct interpretation of a $\mathbf{1\ V}$ source is:
Reasoning: Voltage is defined as energy transferred per unit charge.
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A cell’s $\mathbf{e.m.f.}$ is measured in an open circuit because:
Reasoning: Without current, there is no internal voltage drop, so terminal voltage equals $\mathbf{e.m.f}$.
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If the $\mathbf{e.m.f.}$ of a source is increased, each coulomb of charge receives:
Reasoning: Higher $\mathbf{e.m.f.}$ means greater energy supplied per coulomb.
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The formula $\left( \mathbf{E =}\frac{\mathbf{W}}{\mathbf{Q}} \right)$ indicates that $\mathbf{e.m.f.}$ is a measure of:
Reasoning: $\mathbf{e.m.f.}$ measures energy supplied to each unit of charge.
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A source with zero $\mathbf{e.m.f.}$ would:
Reasoning: Zero $\mathbf{e.m.f.}$ means zero energy transfer per coulomb.
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The complete circuit requirement in the definition of $\mathbf{e.m.f.}$ emphasizes that:
Reasoning: A source drives charge around a complete closed path, restoring charge circulation.
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