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Physics Chapter 11 : Topic 8
Internal Resistance and Terminal Voltage
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The internal resistance of a cell is the resistance offered by:
Reasoning: Internal resistance is the opposition to current flow within the cell itself due to its electrolyte and internal materials.
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When a cell supplies current, its terminal potential difference is:
Reasoning: Some energy is lost inside the cell due to internal resistance, so terminal $\mathbf{p.d.}$ becomes less than $\mathbf{e.m.f.}$
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The relationship between $\mathbf{e.m.f.}$ $\mathbf{(E)}$, terminal voltage $\mathbf{(V)}$, current $\mathbf{(I)}$, and internal resistance $\mathbf{(r)}$ is:
Reasoning: The voltage drop inside the cell is $\mathbf{(Ir)}$, so terminal voltage is $\mathbf{(V = E - Ir)}$.
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A cell has $\mathbf{e.m.f}\mathbf{.}$ $\mathbf{12\ V}$ and internal resistance $\mathbf{1\ \Omega}$. If current supplied is $\mathbf{2\ A}$, the terminal p.d. is:
Reasoning: $\mathbf{V = E - Ir = 12 - (2)(1) = 10\ V}$
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When no current is drawn from a cell, its terminal $\mathbf{p.d.}$ is:
Reasoning: With no current, there is no voltage loss across internal resistance.
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Increasing the current supplied by a cell causes its terminal $\mathbf{p.d.}$ to:
Reasoning: Larger current produces a larger internal voltage drop $\mathbf{(Ir)}$, reducing terminal voltage.
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The terminal potential difference of a cell is:
Reasoning: Terminal $\mathbf{p.d.}$ is the voltage measured across the terminals of a cell when connected to a circuit.
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The energy loss inside a cell due to internal resistance appears mainly as:
Reasoning: Internal resistance converts some electrical energy into heat.
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A battery has high internal resistance. When connected to a heavy load, its terminal voltage will:
Reasoning: High internal resistance causes a large voltage drop $\mathbf{(Ir)}$.
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A practical cell differs from an ideal cell because it has:
Reasoning: Real cells have internal resistance that reduces terminal voltage during operation.
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A cell has $\mathbf{e.m.f.\ \ 6\ V}$ and terminal voltage $\mathbf{5\ V}$ when supplying $\mathbf{2\ A}$ current. Its internal resistance is:
Reasoning: $\mathbf{V = E - Ir}$ $\mathbf{\Longrightarrow}$ So, $\mathbf{r =}$ $\frac{\mathbf{E\ - \ V}}{\mathbf{I}}\mathbf{=}\frac{\mathbf{6\ - \ 5}}{\mathbf{2}}\mathbf{= 0.5\ \Omega}$
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The voltage drop inside a cell is given by:
Reasoning: Current through internal resistance causes a voltage drop equal to $\mathbf{(Ir)}$.
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An ideal cell has internal resistance:
Reasoning: An ideal source has no internal voltage loss, meaning $\mathbf{(r = 0)}$.
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A voltmeter connected across the terminals of a cell measures:
Reasoning: A voltmeter measures the potential difference between the cell terminals.
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A cell delivers maximum terminal voltage when:
Reasoning: With zero current, there is no internal voltage drop, so $\mathbf{(V = E)}$.
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If internal resistance of a cell increases while current remains constant, terminal voltage:
Reasoning: Larger internal resistance increases voltage loss $\mathbf{(Ir)}$.
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The terminal voltage of a cell is lower than its $\mathbf{e.m.f.}$ because:
Reasoning: Internal resistance consumes some energy as heat inside the cell.
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A battery rated at $\mathbf{9\ V}$ gives $\mathbf{8\ V}$ when connected to a circuit. The difference is due to:
Reasoning: The lost $\mathbf{1\ V}$ represents the internal voltage drop.
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The terminal $\mathbf{p.d.}$ can be equal to $\mathbf{e.m.f.}$ when:
Reasoning: No current means no voltage drop across internal resistance.
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The equation $\mathbf{(E = V + Ir)}$ shows that:
Reasoning: The $\mathbf{e.m.f.}$ energy is divided between external circuit and internal resistance.
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A cell with internal resistance $\mathbf{2\ \Omega}$ supplies $\mathbf{3\ A}$ current. The internal voltage loss is:
Reasoning: Voltage loss $\mathbf{= Ir = (3)(2) = 6V}$
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A battery heats up during use because:
Reasoning: Power loss inside the battery is $\mathbf{I}^{\mathbf{2\ }}\mathbf{r}$, which appears as heat.
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To reduce the effect of internal resistance in a battery, one should:
Reasoning: Lower internal resistance reduces voltage loss.
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A cell has $\mathbf{e.m.f.\ 10\ V}$ and internal resistance $\mathbf{0.5\ \Omega}$. If current is $\mathbf{4\ A}$, terminal voltage is:
Reasoning: $\mathbf{V = 10 - (4)(0.5) = 8\ V}$
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A battery’s terminal voltage decreases when supplying more current because:
Reasoning: More current causes a larger $\mathbf{(Ir)}$ drop inside the battery.
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The power wasted inside a cell due to internal resistance is:
Reasoning: Power dissipated by internal resistance is $\mathbf{P =}\mathbf{I}^{\mathbf{2\ }}\mathbf{r}$.
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A cell with very small internal resistance behaves more like:
Reasoning: Small internal resistance causes negligible voltage loss.
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Connecting many cells in series increases the total $\mathbf{e.m.f.}$ but also:
Reasoning: Internal resistances of cells add when connected in series.
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The terminal voltage of a cell is maximum when:
Reasoning: No current means no internal voltage drop.
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Which statement correctly describes internal resistance?
Reasoning: Internal resistance causes an internal voltage drop, reducing terminal $\mathbf{p.d}$.
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A cell’s $\mathbf{e.m.f.}$ remains constant but its terminal voltage decreases during heavy use. The reason is:
Reasoning: Higher current causes greater internal energy loss.
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The terminal voltage of a cell can be greater than its $\mathbf{e.m.f.}$ when:
Reasoning: During charging, an external source can increase the terminal voltage above the cell’s $\mathbf{e.m.f}$.
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A graph of terminal voltage $\mathbf{(V)}$ against current $\mathbf{(I)}$ for a cell is a straight line with:
Reasoning: From $\mathbf{V = E - Ir}$, slope is $\mathbf{( - \ r)}$.
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The $\mathbf{y}$-intercept of a terminal voltage versus current graph represents:
Reasoning: When $\left( \mathbf{I = 0} \right)\mathbf{\ }$, $\mathbf{(V = E)}$.
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A cell is connected to a low-resistance load. The terminal voltage becomes smaller because:
Reasoning: Low external resistance produces high current, increasing $\mathbf{(Ir)}$ loss.
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