Concepts Of Physics MCQ Edition [Volume 2]

Exam: Concepts Of Physics MCQ Edition [Volume 2]
Subject: Physics
Chapter: Electric Current in Conductors

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Electric Current in Conductors

An arrangement designed to measure the emf E and internal...

An arrangement designed to measure the emf

\mathcal{E}

and internal resistance

r

of a battery is depicted in the figure. The voltmeter possesses a very high resistance, whereas the ammeter has some resistance. While the switch

S

is open, the voltmeter indicates

1.52\text{ V}

. Once the switch is closed, the voltmeter's reading decreases to

1.45\text{ V}

and the ammeter displays

1.0\text{ A}

. Calculate the emf and the internal resistance of the battery.
An arrangement designed to measure the emf E and internal resistance r of a battery is depicted in the figure. The

An arrangement designed to measure the emf E and internal resistance r of a battery is depicted in the figure. The

Options

A $1.52\text{ V}$ , $0.07\text{ }\Omega$
B $1.52\text{ V}$ , $0.70\text{ }\Omega$
C $1.45\text{ V}$ , $0.07\text{ }\Omega$
D $1.45\text{ V}$ , $0.70\text{ }\Omega$

Correct answer

A. $1.52\text{ V}$ , $0.07\text{ }\Omega$

Step-by-step solution

When the switch

S

is open, the current in the circuit is zero. The voltmeter measures the emf of the battery.

\mathcal{E} = 1.52\text{ V}

When the switch

S

is closed, a current

I = 1.0\text{ A}

flows through the circuit. The voltmeter measures the terminal voltage

V

across the battery.

The terminal voltage is given by:

V = \mathcal{E} - Ir

Substituting the given values:

1.45 = 1.52 - (1.0)r

r = 1.52 - 1.45 = 0.07\text{ }\Omega

The emf is

1.52\text{ V}

and the internal resistance is

0.07\text{ }\Omega

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