Question

Kunlund 6.00 1 4.00 12 3. (10 pts) A 1.5V battery produces 45 mA when it is across a 30.0 2 load. (a) Find the potential difference across the battery's terminals. (b) What is the battery's internal resistance?

Answer 1

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SOLUTION:

PART (a):

Given that,

Load resistance, $R=30\;\Omega$

Current, $I=45\;mA=45\times10^{-3}\;A$

(1 mA = 10^-3 A)

Therefore, according to Ohm's law, potential difference across the terminals is given by:

$V=I.R=(45\times10^{-3}\;A)\times30\;\Omega$

$\boldsymbol{\therefore V=1.35\;V}$

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PART (b):

We have, EMF of the battery, $E=1.5\;V$

Potential difference across the battery, $V=1.35\;V$

Therefore, Internal resistance of the battery is given by,

$R_{i}=\frac{E-V}{I}=\frac{1.5\;V-1.35\;V}{45\times10^{-3}\;A}$

$\therefore R_{i}=\frac{0.15\;V}{45\times10^{-3}\;A}$

$\boldsymbol{\therefore R_{i}=3.33\;\Omega}$

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