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Byju's Answer

Standard XIII

Physics

Parallel Combination of Cells

A single batt...

Question

A single battery fo emf E and internal resistance r is equivalent to a parallel combination of two batteries of emfs $E_{1} = 2 V$ and $E_{2} = 1.5 V$ and internal resistances $r_{1} = 1 Ω$ and $r_{2} = 2 Ω$ respectively, with polarities as shown in the figure.

(IIT-JEE 1997)

$E = \frac{5}{6} V$

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E = 1.2 V

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r = 1.5 Ω

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$r = \frac{2}{3} Ω$

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Solution

The correct option is A
$E = \frac{5}{6} V$

Simple,
All you have to do is resolve the cell combination.
$E = \frac{E_{1} r_{2} - E_{2} r_{1}}{r_{1} + r_{2}} = \frac{2 \times 2 - 1.5 \times 1}{2 + 1}$
$= \frac{2.5}{3} = \frac{5}{6} V$

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Similar questions

Q. Find the emf (V) and internal resistance (r) of a single battery which is equivalent to a parallel combination of two batteries of emfs

V_{1}

and

V_{2}

and internal resistances

r_{1}

and

r_{2}

respectively, with polarities as shown in figure

Q. The emf (V) and internal resistance (

r

) of a single battery which is equivalent to a parallel combination of two batteries of emf's

V_{1}

and

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and internal resistance

r_{1}

and

r_{2}

, respectively, with polarities as shown in the figure are given by

\frac{V_{1} r_{2} - V_{2} r_{1}}{r_{1} + r_{2}}, \frac{r_{1} r_{2}}{r_{1} + r_{2}}

Q. Two batteries of emf

E_{1}

and

E_{2} (E_{2} > E_{1})

and internal resistance

r_{1}

and

r_{2}

respectively are connected in parallel as shown in figure.

Q. A circuit consists of three batteries of emf

E_{1} = 1 V, E_{2} = 2 V

and

E_{3} = 3 V

and internal resistances

1 Ω, 2 Ω

and

1 Ω

respectively which are connected in parallel as shown in the figure. The potential difference between points

P

and

Q

is:

Obtain the equivalent emf and internal resistance of a parallel combination of 2 cell of emf E1 and E2 and internal resistance R1 and R2.

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A single battery fo emf E and internal resistance r is equivalent to a parallel combination of two batteries of emfs E1=2V and E2=1.5V and internal resistances r1=1Ω and r2=2Ω respectively, with polarities as shown in the figure. (IIT-JEE 1997)

The correct option is A E=56V Simple, All you have to do is resolve the cell combination. E=E1r2−E2r1r1+r2=2×2−1.5×12+1 =2.53=56V

A single battery fo emf E and internal resistance r is equivalent to a parallel combination of two batteries of emfs $E_{1} = 2 V$ and $E_{2} = 1.5 V$ and internal resistances $r_{1} = 1 Ω$ and $r_{2} = 2 Ω$ respectively, with polarities as shown in the figure.

(IIT-JEE 1997)

The correct option is A
$E = \frac{5}{6} V$

Simple,
All you have to do is resolve the cell combination.
$E = \frac{E_{1} r_{2} - E_{2} r_{1}}{r_{1} + r_{2}} = \frac{2 \times 2 - 1.5 \times 1}{2 + 1}$
$= \frac{2.5}{3} = \frac{5}{6} V$