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JAMB Physics 2023 Paper

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Question : 33

Total: 65

A 35 kΩ is connected in series with a resistance of 40 kΩ. What resistance R must be connected in parallel with the combination so that the equivalent resistance is equal to 25 kΩ?

40 kΩ
37.5 kΩ
45.5 kΩ
30 kΩ

Solution:

For the combination in series;

⇒R1 = 35kΩ + 40kΩ = 75kΩ

R is combined with 75kΩ in parallel to give 25kΩ

=

\frac{1}{R_{eq}}

=

\frac{1}{R}

+

\frac{1}{R}

=

\frac{1}{25}

=

\frac{1}{R}

+

\frac{1}{75}

=

\frac{1}{25}

-

\frac{1}{75}

+

\frac{1}{R}

= $\frac{3-1}{75}$ = $\frac{1}{R}$

=

\frac{2}{75}

=

\frac{1}{R}

=

\frac{75}{2}

= R

; R = 37.5k Ω

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