201. Which option correctly compares emf and terminal voltage for a real discharging cell?
ⓐ. \(\mathcal{E}\) is less than \(V\) because internal resistance adds voltage
ⓑ. \(\mathcal{E}\) equals \(V-Ir\)
ⓒ. \(\mathcal{E}\) is always zero when current flows
ⓓ. \(\mathcal{E}\) equals \(V+Ir\)
Correct Answer: \(\mathcal{E}\) equals \(V+Ir\)
Explanation: During discharge, the terminal voltage is \(V=\mathcal{E}-Ir\). Rearranging gives \(\mathcal{E}=V+Ir\). This means the emf is used partly to maintain terminal voltage across the external circuit and partly to overcome internal resistance. The term \(Ir\) represents lost volts inside the cell. This relation also shows why terminal voltage becomes smaller than emf when current is drawn. The sign of \(Ir\) is the main mistake in emf-terminal-voltage comparisons.
202. A \(2\,\text{V}\) cell has internal resistance \(0.2\,\Omega\). It is short-circuited by a wire of negligible resistance. What is the short-circuit current, and why is this condition unsafe?
ⓐ. \(0\,\text{A}\), because external resistance is zero
ⓑ. \(2\,\text{A}\), because current equals emf
ⓒ. \(10\,\text{A}\), limited by internal resistance
ⓓ. \(20\,\text{A}\), because internal resistance is ignored
Correct Answer: \(10\,\text{A}\), limited by internal resistance
Explanation: \( \textbf{Given:} \) \(\mathcal{E}=2\,\text{V}\), \(r=0.2\,\Omega\), and external resistance \(R\approx0\,\Omega\).
\( \textbf{Current relation:} \) \(I=\frac{\mathcal{E}}{R+r}\).
\( \textbf{Short-circuit condition:} \) Since \(R\approx0\), the current is limited mainly by \(r\).
\( \textbf{Substitution:} \) \(I=\frac{2}{0+0.2}\).
\( \textbf{Simplification:} \) \(I=10\,\text{A}\).
\( \textbf{Safety reason:} \) A large current can produce excessive heating inside the cell and connecting wire.
\( \textbf{Final answer:} \) The short-circuit current is \(10\,\text{A}\); zero external resistance does not mean zero current, it means very large current limited only by internal resistance.
203. Two cells of emfs \(\mathcal{E}_1\) and \(\mathcal{E}_2\) with internal resistances \(r_1\) and \(r_2\) are connected in series aiding. Which pair gives the equivalent emf and equivalent internal resistance?
ⓐ. \(\mathcal{E}_{\text{eq}}=\mathcal{E}_1+\mathcal{E}_2\), \(r_{\text{eq}}=r_1+r_2\)
ⓑ. \(\mathcal{E}_{\text{eq}}=\mathcal{E}_1-\mathcal{E}_2\), \(r_{\text{eq}}=r_1-r_2\)
ⓒ. \(\mathcal{E}_{\text{eq}}=\frac{\mathcal{E}_1\mathcal{E}_2}{\mathcal{E}_1+\mathcal{E}_2}\), \(r_{\text{eq}}=\frac{r_1r_2}{r_1+r_2}\)
ⓓ. \(\mathcal{E}_{\text{eq}}=\mathcal{E}_1+\mathcal{E}_2\), \(r_{\text{eq}}=\frac{r_1r_2}{r_1+r_2}\)
Correct Answer: \(\mathcal{E}_{\text{eq}}=\mathcal{E}_1+\mathcal{E}_2\), \(r_{\text{eq}}=r_1+r_2\)
Explanation: In a series-aiding connection, the positive terminal of one cell is connected so that both cells drive current in the same direction around the circuit. Their emfs therefore add algebraically, giving \(\mathcal{E}_{\text{eq}}=\mathcal{E}_1+\mathcal{E}_2\). The internal resistances are also in the same single current path, so they add like series resistors. Hence \(r_{\text{eq}}=r_1+r_2\). The important distinction is that emfs add according to polarity, while internal resistances add in series regardless of which cell has the larger emf.
204. A \(2\,\text{V}\) cell of internal resistance \(0.5\,\Omega\) and a \(3\,\text{V}\) cell of internal resistance \(1.0\,\Omega\) are connected in series aiding with an external resistor of \(3.5\,\Omega\). What current flows?
ⓐ. \(0.80\,\text{A}\)
ⓑ. \(1.00\,\text{A}\)
ⓒ. \(1.25\,\text{A}\)
ⓓ. \(1.43\,\text{A}\)
Correct Answer: \(1.00\,\text{A}\)
Explanation: \( \textbf{Given:} \) \(\mathcal{E}_1=2\,\text{V}\), \(\mathcal{E}_2=3\,\text{V}\), \(r_1=0.5\,\Omega\), \(r_2=1.0\,\Omega\), and \(R=3.5\,\Omega\).
\( \textbf{Series-aiding emf:} \) \(\mathcal{E}_{\text{eq}}=\mathcal{E}_1+\mathcal{E}_2=2+3=5\,\text{V}\).
\( \textbf{Internal resistance:} \) \(r_{\text{eq}}=r_1+r_2=0.5+1.0=1.5\,\Omega\).
\( \textbf{Total circuit resistance:} \) \(R_{\text{total}}=R+r_{\text{eq}}=3.5+1.5=5.0\,\Omega\).
\( \textbf{Current formula:} \) \(I=\frac{\mathcal{E}_{\text{eq}}}{R_{\text{total}}}\).
\( \textbf{Substitution:} \) \(I=\frac{5}{5.0}=1.00\,\text{A}\).
\( \textbf{Final answer:} \) The current is \(1.00\,\text{A}\); the internal resistances must be included because they are part of the complete series path.
205. Two cells of emfs \(6\,\text{V}\) and \(2\,\text{V}\) are connected in series opposing. Their internal resistances are \(1\,\Omega\) and \(1\,\Omega\), respectively. The combination is connected to an external resistor of \(6\,\Omega\). What is the current in the circuit?
ⓐ. \(0.25\,\text{A}\)
ⓑ. \(1.00\,\text{A}\)
ⓒ. \(0.50\,\text{A}\)
ⓓ. \(1.33\,\text{A}\)
Correct Answer: \(0.50\,\text{A}\)
Explanation: \( \textbf{Given:} \) The cell emfs are \(6\,\text{V}\) and \(2\,\text{V}\), internal resistances are \(1\,\Omega\) and \(1\,\Omega\), and external resistance is \(6\,\Omega\).
\( \textbf{Opposing-emf condition:} \) Since the cells oppose each other, the equivalent emf is the difference of emfs.
\( \textbf{Equivalent emf:} \) \(\mathcal{E}_{\text{eq}}=6-2=4\,\text{V}\), in the direction of the stronger cell.
\( \textbf{Internal resistance:} \) The cells are still in series, so \(r_{\text{eq}}=1+1=2\,\Omega\).
\( \textbf{Total resistance:} \) \(R_{\text{total}}=6+2=8\,\Omega\).
\( \textbf{Current:} \) \(I=\frac{\mathcal{E}_{\text{eq}}}{R_{\text{total}}}=\frac{4}{8}=0.50\,\text{A}\).
\( \textbf{Final answer:} \) The current is \(0.50\,\text{A}\); emfs may subtract in opposition, but internal resistances still add.
206. Two identical cells, each of emf \(\mathcal{E}\) and internal resistance \(r\), are connected in series aiding across an external resistance \(R\). Which expression gives the current?
ⓐ. \(I=\frac{\mathcal{E}}{R+2r}\)
ⓑ. \(I=\frac{2\mathcal{E}}{R+r}\)
ⓒ. \(I=\frac{2\mathcal{E}}{R+2r}\)
ⓓ. \(I=\frac{\mathcal{E}}{2R+r}\)
Correct Answer: \(I=\frac{2\mathcal{E}}{R+2r}\)
Explanation: \( \textbf{Series-aiding emf:} \) Two identical cells in series aiding have equivalent emf \(2\mathcal{E}\).
\( \textbf{Series internal resistance:} \) Their internal resistances add, so \(r_{\text{eq}}=2r\).
\( \textbf{Complete circuit resistance:} \) The external resistance \(R\) is in series with the internal resistance \(2r\), giving \(R+2r\).
\( \textbf{Current relation:} \) Current equals equivalent emf divided by total resistance.
\( \textbf{Result:} \) \(I=\frac{2\mathcal{E}}{R+2r}\).
\( \textbf{Final answer:} \) \(I=\frac{2\mathcal{E}}{R+2r}\); series connection increases both emf and internal resistance.
207. For \(n\) identical cells, each of emf \(\mathcal{E}\) and internal resistance \(r\), connected in series aiding with external resistance \(R\), the current is:
ⓐ. \(I=\frac{n\mathcal{E}}{R+nr}\)
ⓑ. \(I=\frac{\mathcal{E}}{R+nr}\)
ⓒ. \(I=\frac{n\mathcal{E}}{nR+r}\)
ⓓ. \(I=\frac{\mathcal{E}}{R+\frac{r}{n}}\)
Correct Answer: \(I=\frac{n\mathcal{E}}{R+nr}\)
Explanation: \( \textbf{Equivalent emf:} \) For \(n\) identical cells connected in series aiding, emfs add to give \(n\mathcal{E}\).
\( \textbf{Equivalent internal resistance:} \) Internal resistances are in series, so they add to \(nr\).
\( \textbf{Total resistance:} \) The external resistance is \(R\), so complete circuit resistance is \(R+nr\).
\( \textbf{Current formula:} \) \(I=\frac{\text{equivalent emf}}{\text{total resistance}}\).
\( \textbf{Substitution:} \) \(I=\frac{n\mathcal{E}}{R+nr}\).
\( \textbf{Final answer:} \) \(I=\frac{n\mathcal{E}}{R+nr}\); the numerator and internal-resistance part both change when cells are added in series.
208. Three identical cells, each of emf \(1.5\,\text{V}\) and internal resistance \(0.2\,\Omega\), are connected in series aiding to an external resistor of \(3.4\,\Omega\). What current flows?
ⓐ. \(0.75\,\text{A}\)
ⓑ. \(1.00\,\text{A}\)
ⓒ. \(1.125\,\text{A}\)
ⓓ. \(1.32\,\text{A}\)
Correct Answer: \(1.125\,\text{A}\)
Explanation: \( \textbf{Given:} \) \(n=3\), \(\mathcal{E}=1.5\,\text{V}\), \(r=0.2\,\Omega\), and \(R=3.4\,\Omega\).
\( \textbf{Equivalent emf:} \) \(\mathcal{E}_{\text{eq}}=n\mathcal{E}=3(1.5)=4.5\,\text{V}\).
\( \textbf{Equivalent internal resistance:} \) \(r_{\text{eq}}=nr=3(0.2)=0.6\,\Omega\).
\( \textbf{Total resistance:} \) \(R_{\text{total}}=R+r_{\text{eq}}=3.4+0.6=4.0\,\Omega\).
\( \textbf{Current:} \) \(I=\frac{\mathcal{E}_{\text{eq}}}{R_{\text{total}}}=\frac{4.5}{4.0}=1.125\,\text{A}\).
\( \textbf{Final answer:} \) The current is \(1.125\,\text{A}\); ignoring internal resistance would give a larger but incorrect current.
209. Two identical cells are connected in series aiding. Compared with a single cell connected to the same external resistance \(R\), the series combination gives nearly double the current only when:
ⓐ. \(R=0\)
ⓑ. \(R\gg r\)
ⓒ. \(R=r\)
ⓓ. \(R\ll r\)
Correct Answer: \(R\gg r\)
Explanation: For a single cell, \(I_1=\frac{\mathcal{E}}{R+r}\). For two identical cells in series aiding, \(I_2=\frac{2\mathcal{E}}{R+2r}\). The current would be nearly doubled only if the internal resistance terms are negligible compared with \(R\). When \(R\gg r\), \(R+r\approx R\) and \(R+2r\approx R\), so \(I_2\approx\frac{2\mathcal{E}}{R}=2I_1\). If \(R\) is comparable to or smaller than \(r\), the added internal resistance significantly reduces the expected gain. A common mistake is to assume series cells always double current.
210. For \(n\) identical cells in series aiding, each with emf \(\mathcal{E}\) and internal resistance \(r\), what is the maximum possible current when the external resistance is negligible?
ⓐ. \(\frac{n\mathcal{E}}{r}\)
ⓑ. \(\frac{\mathcal{E}}{nr}\)
ⓒ. \(\frac{n\mathcal{E}}{R+r}\)
ⓓ. \(\frac{\mathcal{E}}{r}\)
Correct Answer: \(\frac{\mathcal{E}}{r}\)
Explanation: \( \textbf{Series-cell current:} \) For \(n\) identical cells in series, \(I=\frac{n\mathcal{E}}{R+nr}\).
\( \textbf{Negligible external resistance:} \) If \(R\approx0\), the current becomes \(I=\frac{n\mathcal{E}}{nr}\).
\( \textbf{Cancellation:} \) The factor \(n\) cancels from numerator and denominator.
\( \textbf{Maximum current:} \) \(I=\frac{\mathcal{E}}{r}\).
\( \textbf{Physical meaning:} \) Adding identical cells in series increases emf but also increases internal resistance in the same ratio.
\( \textbf{Final answer:} \) The maximum short-circuit current remains \(\frac{\mathcal{E}}{r}\) for identical series cells; this shows why internal resistance limits current strongly.
211. A \(4\,\text{V}\) cell of internal resistance \(1\,\Omega\) is connected in series opposing with a \(1.5\,\text{V}\) cell of internal resistance \(0.5\,\Omega\). The combination is connected to \(3.5\,\Omega\). Which direction and magnitude of current are correct?
ⓐ. \(0.50\,\text{A}\), in the direction of the \(4\,\text{V}\) cell
ⓑ. \(0.50\,\text{A}\), in the direction of the \(1.5\,\text{V}\) cell
ⓒ. \(1.10\,\text{A}\), in the direction of the \(4\,\text{V}\) cell
ⓓ. \(1.10\,\text{A}\), in the direction of the \(1.5\,\text{V}\) cell
Correct Answer: \(0.50\,\text{A}\), in the direction of the \(4\,\text{V}\) cell
Explanation: \( \textbf{Opposing cells:} \) The effective emf is the difference between the two emfs.
\( \textbf{Equivalent emf:} \) \(\mathcal{E}_{\text{eq}}=4.0-1.5=2.5\,\text{V}\).
\( \textbf{Direction:} \) The stronger \(4\,\text{V}\) cell decides the direction of current.
\( \textbf{Total internal resistance:} \) \(r_{\text{eq}}=1.0+0.5=1.5\,\Omega\).
\( \textbf{Total circuit resistance:} \) \(R_{\text{total}}=3.5+1.5=5.0\,\Omega\).
\( \textbf{Current:} \) \(I=\frac{2.5}{5.0}=0.50\,\text{A}\).
\( \textbf{Final answer:} \) The current is \(0.50\,\text{A}\) in the direction of the \(4\,\text{V}\) cell; internal resistances add even when emfs oppose.
212. Consider the following statements about cells in series.
Statement I: In series aiding, emfs add.
Statement II: In series opposing, internal resistances subtract.
Statement III: In series connection, the same current passes through each cell.
Which statements are correct?
ⓐ. I and II only
ⓑ. I and III only
ⓒ. II and III only
ⓓ. I, II and III
Correct Answer: I and III only
Explanation: Statement I is correct because cells in series aiding drive current in the same direction, so their emfs add. Statement II is incorrect because internal resistances do not subtract in a series path; they add as ordinary resistances. Statement III is correct because a series circuit has only one path, so the same current passes through every cell. In opposing connection, only the emfs are algebraically subtracted according to polarity. The common sign-related mistake is to subtract internal resistance along with opposing emf.
213. In a series combination of cells, why is the arrangement not always better than using one cell?
ⓐ. Series cells always reduce total emf
ⓑ. Series cells make current zero in every case
ⓒ. Internal resistance disappears in series
ⓓ. Internal resistance also increases in series
Correct Answer: Internal resistance also increases in series
Explanation: Series aiding cells increase the total emf, which can help drive a larger current through a large external resistance. However, their internal resistances also add. If the external resistance is small or comparable to the total internal resistance, the added internal resistance can significantly limit the current. Therefore the benefit of series connection depends on the load resistance. The practical rule is that series cells are most useful when a higher voltage is needed and the external resistance is relatively large.
214. Identical cells are connected in parallel with the same polarity. If each cell has emf \(\mathcal{E}\) and internal resistance \(r\), what are the equivalent emf and internal resistance for \(n\) cells?
ⓐ. \(\mathcal{E}_{\text{eq}}=n\mathcal{E}\), \(r_{\text{eq}}=nr\)
ⓑ. \(\mathcal{E}_{\text{eq}}=\mathcal{E}\), \(r_{\text{eq}}=\frac{r}{n}\)
ⓒ. \(\mathcal{E}_{\text{eq}}=\frac{\mathcal{E}}{n}\), \(r_{\text{eq}}=nr\)
ⓓ. \(\mathcal{E}_{\text{eq}}=n\mathcal{E}\), \(r_{\text{eq}}=\frac{r}{n}\)
Correct Answer: \(\mathcal{E}_{\text{eq}}=\mathcal{E}\), \(r_{\text{eq}}=\frac{r}{n}\)
Explanation: Identical cells connected in parallel with the same polarity maintain the same emf as a single cell. The advantage is not increased emf, but reduced effective internal resistance. Since the internal resistances are effectively in parallel for identical cells, the equivalent internal resistance becomes \(\frac{r}{n}\). This allows the combination to supply a larger current to low external resistance than one cell can. A common mistake is to add emfs in a parallel connection as if the cells were in series.
215. Three identical cells, each of emf \(1.5\,\text{V}\) and internal resistance \(0.6\,\Omega\), are connected in parallel with the same polarity. The combination is connected to an external resistance of \(1.3\,\Omega\). What current flows?
ⓐ. \(0.50\,\text{A}\)
ⓑ. \(1.50\,\text{A}\)
ⓒ. \(3.00\,\text{A}\)
ⓓ. \(1.00\,\text{A}\)
Correct Answer: \(1.00\,\text{A}\)
Explanation: \( \textbf{Given:} \) \(n=3\), \(\mathcal{E}=1.5\,\text{V}\), \(r=0.6\,\Omega\), and \(R=1.3\,\Omega\).
\( \textbf{Parallel-cell emf:} \) For identical cells in parallel, \(\mathcal{E}_{\text{eq}}=\mathcal{E}=1.5\,\text{V}\).
\( \textbf{Equivalent internal resistance:} \) \(r_{\text{eq}}=\frac{r}{n}=\frac{0.6}{3}=0.2\,\Omega\).
\( \textbf{Total resistance:} \) \(R_{\text{total}}=R+r_{\text{eq}}=1.3+0.2=1.5\,\Omega\).
\( \textbf{Current:} \) \(I=\frac{\mathcal{E}_{\text{eq}}}{R_{\text{total}}}=\frac{1.5}{1.5}=1.00\,\text{A}\).
\( \textbf{Final answer:} \) The current is \(1.00\,\text{A}\); parallel cells reduce internal resistance but do not multiply emf.
216. For \(n\) identical cells connected in parallel to an external resistance \(R\), each cell has emf \(\mathcal{E}\) and internal resistance \(r\). Which expression gives the current in the external circuit?
ⓐ. \(I=\frac{n\mathcal{E}}{R+nr}\)
ⓑ. \(I=\frac{\mathcal{E}}{R+\frac{r}{n}}\)
ⓒ. \(I=\frac{\mathcal{E}}{nR+r}\)
ⓓ. \(I=\frac{n\mathcal{E}}{R+\frac{r}{n}}\)
Correct Answer: \(I=\frac{\mathcal{E}}{R+\frac{r}{n}}\)
Explanation: \( \textbf{Parallel-cell emf:} \) Identical cells connected in parallel have equivalent emf \(\mathcal{E}\).
\( \textbf{Parallel internal resistance:} \) Their effective internal resistance is \(\frac{r}{n}\).
\( \textbf{Complete circuit resistance:} \) The external resistance is \(R\), so total resistance is \(R+\frac{r}{n}\).
\( \textbf{Current relation:} \) Current is equivalent emf divided by complete circuit resistance.
\( \textbf{Result:} \) \(I=\frac{\mathcal{E}}{R+\frac{r}{n}}\).
\( \textbf{Final answer:} \) \(I=\frac{\mathcal{E}}{R+\frac{r}{n}}\); this differs from series cells, where both emf and internal resistance are multiplied by \(n\).
217. Two identical cells, each of emf \(2\,\text{V}\) and internal resistance \(1\,\Omega\), are connected in parallel to a \(1\,\Omega\) external resistor. What is the current through the external resistor?
ⓐ. \(0.67\,\text{A}\)
ⓑ. \(1.0\,\text{A}\)
ⓒ. \(1.33\,\text{A}\)
ⓓ. \(2.0\,\text{A}\)
Correct Answer: \(1.33\,\text{A}\)
Explanation: \( \textbf{Given:} \) Two identical cells have \(\mathcal{E}=2\,\text{V}\), \(r=1\,\Omega\), and external resistance \(R=1\,\Omega\).
\( \textbf{Equivalent emf:} \) For identical cells in parallel, \(\mathcal{E}_{\text{eq}}=2\,\text{V}\).
\( \textbf{Equivalent internal resistance:} \) \(r_{\text{eq}}=\frac{r}{2}=\frac{1}{2}=0.5\,\Omega\).
\( \textbf{Total resistance:} \) \(R_{\text{total}}=R+r_{\text{eq}}=1+0.5=1.5\,\Omega\).
\( \textbf{Current:} \) \(I=\frac{2}{1.5}=1.33\,\text{A}\) approximately.
\( \textbf{Final answer:} \) The current is \(1.33\,\text{A}\); the emf is not \(4\,\text{V}\) because the cells are in parallel, not series.
218. Which statement best explains why identical cells are often connected in parallel?
ⓐ. To increase the emf in proportion to the number of cells
ⓑ. To reduce internal resistance and share current
ⓒ. To make the terminal voltage always zero
ⓓ. To make each cell oppose the others
Correct Answer: To reduce internal resistance and share current
Explanation: Identical cells in parallel have the same equivalent emf as one cell. Their main advantage is that the effective internal resistance is reduced to \(\frac{r}{n}\). This allows the combination to supply more current to a low-resistance load without each cell providing the entire load current alone. The load current is shared among identical cells. This arrangement is useful when current capacity is important rather than higher voltage. The common misconception is to expect the emf to add in parallel.
219. Two identical cells are available, each with emf \(\mathcal{E}\) and internal resistance \(r\). For a very large external resistance \(R\), which connection gives the larger current approximately?
ⓐ. Series, because emf nearly doubles when \(R\) is very large
ⓑ. Parallel, because the effective internal resistance becomes zero
ⓒ. Both give exactly the same current
ⓓ. Neither gives current when \(R\) is large
Correct Answer: Series, because emf nearly doubles when \(R\) is very large
Explanation: For two cells in series, \(I_s=\frac{2\mathcal{E}}{R+2r}\). For two identical cells in parallel, \(I_p=\frac{\mathcal{E}}{R+\frac{r}{2}}\). If \(R\) is very large compared with \(r\), then \(R+2r\approx R\) and \(R+\frac{r}{2}\approx R\). Thus \(I_s\approx\frac{2\mathcal{E}}{R}\), while \(I_p\approx\frac{\mathcal{E}}{R}\). Series gives nearly twice the current in this high-resistance-load limit. The condition \(R\gg r\) is essential because internal resistance becomes less important.
220. Two identical cells are available, each with emf \(\mathcal{E}\) and internal resistance \(r\). For a very small external resistance \(R\), which connection is generally better for supplying larger external current?
ⓐ. Series, because it doubles emf without changing internal resistance
ⓑ. Both always supply the same current for every \(R\)
ⓒ. Opposing series, because it cancels internal resistance
ⓓ. Parallel, because it reduces effective internal resistance
Correct Answer: Parallel, because it reduces effective internal resistance
Explanation: For two cells in series, \(I_s=\frac{2\mathcal{E}}{R+2r}\). For two identical cells in parallel, \(I_p=\frac{\mathcal{E}}{R+\frac{r}{2}}\). If \(R\) is very small compared with \(r\), then internal resistance strongly controls the current. In the limiting case \(R\approx0\), series gives \(I_s\approx\frac{2\mathcal{E}}{2r}=\frac{\mathcal{E}}{r}\), while parallel gives \(I_p\approx\frac{\mathcal{E}}{r/2}=\frac{2\mathcal{E}}{r}\). Parallel is better for low-resistance loads because it reduces the internal resistance. The practical choice between series and parallel depends on whether higher voltage or lower internal resistance is needed.
