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WAEC Physics 2022 Paper

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Question : 25

Total: 50

A galvanometer with a full-scale deflection of 20 mA is Converted to read 8 K by connecting a

395\ \Omega

resistor in series with it. Determine the internal resistance of the galvanometer

$2.5\ \Omega$
$5.0\ \Omega$
$8.0\ \Omega$
$10.0\ \Omega$

Solution:

r=

resistance of the galvanometer

I=

current through galvanometer

=20\ \mathrm{mA}

or 0.02 A

V_1 = p \cdot d

across the galvanometer

= I X_r = 0.02 X_r

: V_1 = 0.02 r

V_2 = p \cdot d

across the multiplier

= 8 - 0.02 r

R =

resistance of the multiplier

=395\ \Omega

where

R = \frac{v}{I}

\rightarrow 395 = \frac{8 - 0.02 r}{0.02}

CROSS MULTIPLY

\; 8 - 0.02 r = 395^* 0.02

\; 8 - 0.02 r = 7.9

\; \rightarrow 0.02 r = 8 - 7.9

\; \therefore r = \frac{0.1}{0.02}

\; = 5\ \Omega

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