Cells in Series and in Parallel MCQ Quiz - Objective Question with Answer for Cells in Series and in Parallel - Download Free PDF

The correct answer is: Option 1) √R₁R₂

Concepts:

The internal resistance r of the cell can be determined by comparing the power delivered in two different cases.

In both cases, the power delivered is the same, and the current through the circuit is determined by the external resistance and the internal resistance of the cell.

Given the relationship between current and resistance, we use the formula for power: P = I²R.

Calculation:
Let the current provided by the cell be:

I = E / (R + r), where E is the EMF of the cell, R is the external resistance, and r is the internal resistance of the cell.

For the second case, where the cell is connected to a resistance R₂, the power delivered is:

P₂ = I²R₂ = (E / (R₂ + r))² R₂

The power delivered in both cases is the same, so we equate the powers in both cases:

(E / (R₂ + r))² R₁ = (E / (R₂ + r))² R₂

R₁ (R₂² + r² + 2R₂r) = R₂ (R₁²r² + r²2R₁r)

Solving gives: r = √(R₁R₂)

Concept:

The total resistance in the circuit with a single cell is the sum of the internal resistance of the cell and the resistance of the wire. The current through the wire can be calculated using Ohm's law.

Explanation:

Initially, the current through the wire with a single cell is given by:

I = E / (R + r)

where,

• E = 1.1V (emf of the cell)
• R = 0.5Ω (resistance of the wire)
• r = 0.5Ω (internal resistance of the cell)

So, the current I is:

I = 1.1V / (0.5Ω + 0.5Ω) = 1.1V / 1.0Ω = 1.1A

When another cell of the same emf is connected in series, the total emf becomes:

2E = 2 × 1.1V = 2.2V

Let the internal resistance of the second cell be r₂. The total resistance in the circuit now is:

R + r + r₂

According to the question, the current remains the same, i.e., 1.1A.

So, using Ohm's law again:

1.1A = 2.2V / (R + r + r₂)

Substituting the values:

1.1A = 2.2V / (0.5Ω + 0.5Ω + r₂)

1.1A = 2.2V / (1.0Ω + r₂)

Solving for r₂:

1.0Ω + r₂ = 2.2V / 1.1A

1.0Ω + r₂ = 2.0Ω

r₂ = 2.0Ω - 1.0Ω

r₂ = 1.0Ω

The correct answer is option 1.

Concept:

When two cells are connected in parallel, the net electromotive force (E_eq) and the equivalent internal resistance (r_eq) are calculated using the formulas:

E_eq = (E₁ × r₂ + E₂ × r₁) / (r₁ + r₂)

r_eq = (r₁ × r₂) / (r₁ + r₂)

Here, E₁ and E₂ are the emfs of the two cells, and r₁ and r₂ are their internal resistances respectively.

Explanation:

Given:

• Cell 1: E₁ = 2 V, r₁ = 1 Ω
• Cell 2: E₂ = 1 V, r₂ = 2 Ω
• External resistance R = 4/3 Ω

Step 1: Calculate E_eq and r_eq

E_eq = (2 × 2 + 1 × 1) / (1 + 2) = (4 + 1) / 3 = 5 / 3 V

r_eq = (1 × 2) / (1 + 2) = 2 / 3 Ω

Step 2: Use Ohm's law to calculate current (I)

Total resistance = R + r_eq = (4/3 + 2/3) = 6/3 = 2 Ω

I = E_eq / (R + r_eq) = (5/3) / 2 = 5 / 6 A

The correct answer is option 4.

CONCEPT:

Cell:

• The cell converts chemical energy into electrical energy.
• Cells are of two types:
  1. Primary cell: This type of cell cannot be recharged.
  2. Secondary cell: This type of cell can be recharged.
• For a cell of emf E and internal resistance r,

⇒ E - V = Ir

Where I = current, and V = potential difference across external resistance

Cells in series:

• If the number of cells are connected end to end that the positive terminal of one cell is connected to the negative terminal of the succeeding cell then it is called a series arrangement of cells.
• The equivalent emf of cells in series arrangement is given as,

⇒ Eeq = E1 + E2 +...+ En

• The equivalent internal resistance of cells in a series arrangement is given as,

⇒ req = r1 + r2 +...+ rn

Cells in parallel:

• If the number of cells is connected such that the positive terminals are connected together at one point and the negative terminals of these cells are connected together at another point then it is called a parallel arrangement of cells.
• The equivalent emf of cells in a parallel arrangement is given as,

\(⇒ E_{eq}=\frac{E_1/r_1+E_2/r_2+E_3/r_3+...+E_n/r_n}{1/r_1+1/r_2+1/r_3+...+1/r_n}\)

• The equivalent internal resistance of cells in a parallel arrangement is given as,

\(⇒ \frac{1}{r_{eq}}=\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}+...+\frac{1}{r_n}\)

Calculation:

= \(\rm \frac{\frac{\varepsilon_{1}}{r_{1}}+\frac{\varepsilon_{2}}{r_{2}}}{\frac{1}{r_{1}}+\frac{1}{r_{2}}}=\varepsilon\)

Calculation:

Let \(n\) be the number of wrongly connected cells.

Number of cells helping one another \((12-n)\)

Total e.m.f. of such cells \((12-n)E\)

Total e.m.f. of cells opposing \(nE\)

Resultant e.m.f of battery

\((12-n)E-nE=(12-2n)E\)

Total resistance of cells \(=12r\)

\(\because \) Resistance remains same irrespective of connection of cells 

With additional cells

(a) Total e.m.f. of cells when additional cells help battery

\((12-2n)E+2E\)

Total resistance \(=12r+2r=14r\)

\(\therefore \dfrac {(12-2n)E+2E}{14r}=3\)                ......(i)

(b) similarly when additional cells oppose the battery

\(\dfrac{(12-2n)E-2E}{14r}=2\)                    ....(ii)

Solving (i) and (ii), we will get

\(n=1\)

CONCEPT:

• Cell: An electric cell is a device, which converts chemical energy into electrical energy.
• It has two terminals, which are made up of metal: one terminal is positive, while the other one is negative. When the two terminals are connected to an electrical device, electric current flows through it.
• Battery: A combination of two or more cells electrically connected to work together to produce electric energy.

EXPLANATION:

From the above discussion, we can say that the symbol of the battery is: 

• The first option is a symbol of an earthling. So it does not follow.
• The second option is a symbol of a battery. So it follows.
• The third option is a symbol of a capacitor. So it does not follow.

So option 2 is correct.

 It represents earthing.

 It shows a capacitor.

CONCEPT:

• Grouping of Cells: In a series grouping of cell’s their emf’s are additive or subtractive while their internal resistances are always additive.
• Series grouping: In series grouping anode of one cell is connected to the cathode of other cells and so on. 

• If n identical cells are connected in series

1. Equivalent emf of the combination  E_eq = nE
2. Equivalent internal resistance r_eq = nr

• Parallel grouping: In parallel grouping, all anodes are connected at one point and all cathode are connected together at other points. 

• If n identical cells are connected in parallel

1. Equivalent emf of the combination  E_eq = E
2. Equivalent internal resistance r_eq = r/n

EXPLANATION:

• Therefore, from above it is clear that the potential difference will be maximum in series combination and internal resistance will be minimum in parallel combination.
• The both the emf and internal resistances will be maximum in case of series combination. So option 4 is correct.
• In case of parallel combination, the equivalent emf remains same.

CONCEPT:

• Grouping of Cells: In a series grouping of cell’s their emf’s are additive or subtractive while their internal resistances are always additive.
• If dissimilar plates of cells are connected together then their emf’s are added to each other while if their similar plates are connected together then their emf’s are subtractive

Series grouping: 

• In series grouping anode of one cell is connected to the cathode of other cells and so on.

• If n identical cells are connected in series

1. Equivalent emf of the combination Eeq = nE
2. Equivalent internal resistance req = nr

Parallel grouping:

• In parallel grouping, all anodes are connected at one point and all cathode are connected together at other points.

• If n identical cells are connected in parallel

1. Equivalent emf of the combination Eeq = E
2. Equivalent internal resistance req = r/n

CALCULATION:

Given - number of cell = 5 and internal resistances (r) = 1.5 Ω

• The equivalent internal resistance of the cells is 

⇒ req = r/n

⇒ req = 1.5/5 = 0.3 Ω 

CONCEPT:

• Grouping of Cells: In a series grouping of cell’s their emf’s are additive or subtractive while their internal resistances are always additive.
• Series grouping: In series grouping anode of one cell is connected to the cathode of other cells and so on. 

• If n identical cells are connected in series

1. Equivalent emf of the combination  E_eq = nE
2. Equivalent internal resistance r_eq = nr

CALCULATION:

Given- emf of three cells = E₁ = E₂ = E₃ = E and internal resistance = r₁ = r₂ = r₃ = r

Equivalent emf of the combination  is

⇒ E_eq = E₁ + E₂ + E₃ = E + E + E = 3E

Similarly,

Equivalent internal resistance  is

⇒ r_eq = r₁ + r₂ + r₃ = r + r + r = 3r

So option 3 is correct.

CONCEPT:

• Grouping of Cells: In a series grouping of cell’s their emf’s are additive or subtractive while their internal resistances are always additive.
• If dissimilar plates of cells are connected together then their emf’s are added to each other while if their similar plates are connected together then their emf’s are subtractive

EXPLANATION:

• Series grouping: In series grouping anode of one cell is connected to the cathode of other cells and so on.

• If n identical cells are connected in series

1. Equivalent emf of the combination E_eq = nE
2. Equivalent internal resistance r_eq = nr

• Parallel grouping: In parallel grouping, all anodes are connected at one point and all cathode are connected together at other points.

• If n identical cells are connected in parallel

1. Equivalent emf of the combination E_eq = E
2. Equivalent internal resistance r_eq = r/n

• Therefore, from above it is clear that the potential difference will be maximum in series combination and internal resistance will be minimum in parallel combination. Thus option 4 is correct.

Concept:

• Grouping of Cells: In a series grouping of cell’s their emf’s are additive or subtractive while their internal resistances are always additive.
• If dissimilar plates of cells are connected together then their emf’s are added to each other while if their similar plates are connected together then their emf’s are subtractive

• Series grouping: In series grouping anode of one cell is connected to the cathode of other cells and so on.

• If n identical cells are connected in series

1. Equivalent emf of the combination Eeq = nE
2. Equivalent internal resistance req = nr

• Parallel grouping: In parallel grouping, all anodes are connected at one point and all cathode are connected together at other points.

• If n identical cells are connected in parallel

1. Equivalent emf of the combination Eeq = E
2. Equivalent internal resistance req = r/n

• Kirchoff Voltage Law: The net sum of potential difference across the loop is zero.

Calculation:

• Here it is given that, the two Cells are in Parallel and no other information is given. 
• In this case we can assume that the internal resistances are zero. 
• So, we can simply apply the Kirchoff Voltage Law in the loop.

V_eff = E₁ - E₂

(Given r₁ = r₂ = 0)

Given the EMF of one cell is E1 = 1.25 V

The EMF of the other cell is  E2 = 0.75 V

Veff = E1 - E2 = 1.25 V - 0.75 V = 0.5V

So, the correct option is 0.5 V

Confusion Point:

• Internal resistance is zero when not given.  

CONCEPT:

• Grouping of Cells: In a series grouping of cell’s their emf’s are additive or subtractive while their internal resistances are always additive.
• If dissimilar plates of cells are connected together then their emf’s are added to each other while if their similar plates are connected together then their emf’s are subtractive

EXPLANATION:

• Series grouping: In series grouping anode of one cell is connected to the cathode of other cells and so on.

• If n identical cells are connected in series

1. Equivalent emf of the combination E_eq = nE
2. Equivalent internal resistance r_eq = nr

• Parallel grouping: In parallel grouping, all anodes are connected at one point and all cathode are connected together at other points.

• If n identical cells are connected in parallel

1. Equivalent emf of the combination E_eq = E
2. Equivalent internal resistance r_eq = r/n

• Therefore, from above it is clear that the equivalent emf is the same in the parallel combination of cells. Therefore statement 2 is correct.

CONCEPT:

Cell:

• The cell converts chemical energy into electrical energy.
• Cells are of two types:
  1. Primary cell: This type of cell cannot be recharged.
  2. Secondary cell: This type of cell can be recharged.
• For a cell of emf E and internal resistance r,

⇒ E - V = Ir

Where I = current, and V = potential difference across external resistance

Cells in series:

• If the number of cells are connected end to end that the positive terminal of one cell is connected to the negative terminal of the succeeding cell then it is called a series arrangement of cells.
• The equivalent emf of cells in series arrangement is given as,

⇒ Eeq = E1 + E2 +...+ En

• The equivalent internal resistance of cells in a series arrangement is given as,

⇒ req = r1 + r2 +...+ rn

Cells in parallel:

• If the number of cells is connected such that the positive terminals are connected together at one point and the negative terminals of these cells are connected together at another point then it is called a parallel arrangement of cells.
• The equivalent emf of cells in a parallel arrangement is given as,

\(⇒ E_{eq}=\frac{E_1/r_1+E_2/r_2+E_3/r_3+...+E_n/r_n}{1/r_1+1/r_2+1/r_3+...+1/r_n}\)

• The equivalent internal resistance of cells in a parallel arrangement is given as,

\(⇒ \frac{1}{r_{eq}}=\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}+...+\frac{1}{r_n}\)

EXPLANATION:

• We know that the equivalent emf of cells in a series arrangement is given as,

⇒ Eeq = E1 + E2 +...+ En       ----(1)

• We know that the equivalent emf of cells in a parallel arrangement is given as,

\(⇒ E_{eq}=\frac{E_1/r_1+E_2/r_2+E_3/r_3+...+E_n/r_n}{1/r_1+1/r_2+1/r_3+...+1/r_n}\)       ----(2)

• Bu equation 1 and equation 2, it is clear that the equivalent emf will be more in a series arrangement. Hence, option 1 is correct.

Explanation:

• Primary cells are the cells that produce electrical energy and after a few hours of use become inactive or useless
• Types of a primary cell are: Voltaic cell, Daniel Cell, Lechlanche cell, dry cell
• Secondary cells are used repeatedly for producing electrical energy after charging them.
• The secondary cell does not produce electrical energy, but it only accumulates the energy
• Types of secondary cell: Lead Acid, Edison cell, Nickel Cadmium cell
• A solar cell, or photo-voltaic cell, is an electrical device that converts the energy of light directly into electricity by the photo-voltaic effect.
• Solar cells are the building blocks of photo-voltaic modules, otherwise known as solar panels.
• And since Nickel cadmium cell can't produce electricity on its own, it is used as rechargeable  battery which is classified as secondary cell 

CONCEPT:

Cell:

• The cell converts chemical energy into electrical energy.
• Cells are of two types:
  1. Primary cell: This type of cell cannot be recharged.
  2. Secondary cell: This type of cell can be recharged.
• For a cell of emf E and internal resistance r,

⇒ E - V = Ir

Where I = current, and V = potential difference across external resistance

Cells in series:

• If the number of cells is connected end to end then it is called a series arrangement of cells.
• The equivalent emf of cells in series arrangement is given as,

⇒ E_eq = E₁ + E2 +...+ En

• The equivalent internal resistance of cells in a series arrangement is given as,

⇒ req = r1 + r2 +...+ rn

CALCULATION:

Given:

E₁ = E₂ = 7V, r₁ = r₂ = 1Ω, and R = 5Ω

• The two cells are connected in series so the equivalent emf is given as,

⇒ Eeq = E1 + E2

⇒ Eeq = 7 + 7

⇒ Eeq = 14 V     -----(1)

• And the equivalent resistance is given as,

⇒ req = r1 + r2

⇒ req = 1 + 1

⇒ req = 2 Ω

• The equivalent resistance of the circuit is given as,

⇒ R_eq = R + req

⇒ Req = 5 + 2

⇒ Req = 7 Ω     -----(2)

By Ohm's law,

\(⇒ I=\frac{V}{R_{eq}}\)

\(⇒ I=\frac{14}{7}\)

⇒ I = 2 A

• Hence, option 1 is correct.