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When cells are connected in series (A) the EMF increases. (B) the potential difference decreases. (C) the current capacity increases. (D) the current capacity decreases.

Short Answer

Expert verified
When cells are connected in series, the EMF increases because the voltages across the cells add up while the current capacity remains the same. So, the correct statement is (A) the EMF increases.

Step by step solution

01

Recall Connection in Series#

When two or more cells are connected in series, their positive terminals are connected to the negative terminals of the next cell. In this configuration, the voltages across the cells add up while the current capacity remains the same. Now let's analyze the given options one by one.
02

Option A: EMF increases#

When cells are connected in series their EMFs (Electromotive force) are added up. In this case, the effective EMF becomes the sum of all the individual cells' EMFs. So the statement (A) the EMF increases, is correct.
03

Option B: Potential difference decreases#

The potential difference across the cells in series is equal to the sum of the potential difference across each cell. So, when connected in series, the potential difference does not decrease. Therefore, this statement is false.
04

Option C: Current capacity increases#

When cells are connected in series, the current capacity remains the same. The arrangement does not increase the current capacity. So, this statement is false.
05

Option D: Current capacity decreases#

As mentioned earlier, when cells are connected in series, their current capacity remains the same. It does not decrease. Thus, this statement is also false. Among all the given options, only Option A (the EMF increases) correctly describes what happens when cells are connected in series.

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Most popular questions from this chapter

What is the current through the resistor \(R\) in the circuit shown below? The EMF of each cell is \(E_{m}\) and internal resistance is \(r\) (A) \(\frac{E_{m}}{2 R+r}\) (B) \(\frac{E_{m}}{2 r+R}\) (C) \(\frac{2 E_{m}}{R+2 r}\) (D) \(\frac{2 E_{m}}{2 R+r}\)

An electric current is passed through a circuit containing two wires of the same material, connected in parallel. If the lengths and radii of the wires are in the ratio of \(4 / 3\) and \(2 / 3\), then the ratio of the currents passing through the wire will be (A) 3 (B) \(1 / 3\) (C) \(\underline{8 / 9}\) (D) 2

Figure \(14.47\) shows the circuit of a potentiometer. The length of the potentiometer wire \(A B\) is \(50 \mathrm{~cm}\). The EMF \(E_{1}\) of the battery is \(4 \mathrm{~V}\), having negligible internal resistance. Value of \(R_{1}\) and \(R_{2}\) are \(15 \Omega\) and \(5 \Omega\), respectively. When both the keys are open, the null point is obtained at a distance of \(31.25 \mathrm{~cm}\) from \(A\), but when both the keys are closed, the balance length reduces to \(5 \mathrm{~cm}\) only. Given \(R_{A B}=10 \Omega\) The internal resistance of the cell \(E_{2}\) is (A) \(4.5 \Omega\) (B) \(5.5 \Omega\) (C) \(6.5 \Omega\) (D) \(7.5 \Omega\)

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A wire with resistance \(12 \Omega\) is bent in the form of a circle. The effective resistance between the two points on any diameter of the circle is (A) \(12 \Omega\) (B) \(24 \Omega\) (C) \(6 \Omega\) (D) \(3 \Omega\)

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