A cell of emf 4 ~V and internal resistance 0.5 is connected to a 7.5 external resistance. The terminal potential difference of the cell is:

Correct answer: A. 3.75 ~V As, i= E R+r Hence, i=(4)/(7.5+0.5)=(4)/(8)=(1)/(2) And, V=E-i r (where, the symbols have their usual meanings) aligned V & =4-0.5 × 0.5 V & =3.75 volt aligned

NEET 2022

Exam: NEET
Year: 2022
Subject: Physics
Chapter: Current Electricity

NEET 2022 · Physics · Current Electricity

A cell of emf 4 ~V and internal resistance 0.5 is connected to a...

A cell of emf

4 \mathrm{~V}

and internal resistance

0.5 \Omega

is connected to a

7.5 \Omega

external resistance. The terminal potential difference of the cell is:

Options

A $3.75 \mathrm{~V}$
B $4.25 \mathrm{~V}$
C $4 \mathrm{~V}$
D $0.375 \mathrm{~V}$

Correct answer

A. $3.75 \mathrm{~V}$

Step-by-step solution

As,

i=\frac{\mathrm{E}}{\mathrm{R}+r}

Hence,

\quad i=\frac{4}{7.5+0.5}=\frac{4}{8}=\frac{1}{2}

And,

\mathrm{V}=\mathrm{E}-i r

(where, the symbols have their usual meanings)

\begin{aligned} \mathrm{V} & =4-0.5 \times 0.5 \\ \mathrm{ V} & =3.75 \text{ volt } \end{aligned}

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