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2017 WAEC Physics Theory  (i) What is dielectric? (ii) A parallel plate capacitor consists of two plates each of...

Physics
WAEC 2017

 (i) What is dielectric?

(ii) A parallel plate capacitor consists of two plates each of area 9.6 x 10\(^{-2}\)m\(^2\), separated by a dielectric of thickness 2.25 x 10\(^{-3}\) and dielectric constant 900. Calculate the capacitance of the capacitor. [ \(\varepsilon_o\) = permittivity of free space = 8.85 x 10\(^{-12}\)Fm\(^{-1}\)]

(b) (i) Which of the following devices has a higher resistance; an ammeter or a voltmeter? Give a reason for your answer.

(ii)


The resistance of the voltmeter in the circuit diagram illustrated above is 800\(\Omega\), Calculate the voltmeter reading.

(c)


A battery of negligible internal resistance is connected to a set of resistors as illustrated in the circuit diagram above. Determine the equivalent resistance of the circuit.


 

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Explanation

(a)(i) A dielectric is an insulating material (solid, liquid or gas) separating the two plates of a capacitor.

(ii) Using C = \(\frac{\xi_1.\xi_oA}{d} = \frac{900 \times 8.85 \times 10^{-12} \times 9.6 \times 10^{-3}}{2.25 \times 10^{-4}}\)

= 3.4 x 10\(^{-7}\)

 

 (b)(i) A voltmeter has a higher resistance. Reason: A voltmeter is connected inparallel in a circuit to allow passage of minimal current through it; hence the circuit current is not significantly affected.

(ii)

\(\frac{1}{R_1} = \frac{1}{400} + \frac{1}{400}\)    \(R_1 = 200\Omega\)

\(\frac{1}{R_2} = \frac{1}{800} + \frac{1}{800}\)    \(R_1 = 400\Omega\)

RT= 200 + 400 = 600\(\Omega\)

I = \(\frac{V_T}{R_T} = \frac{6}{600}\) = 0.01A

Voltmeter reading

= Pd across R\(_2\)

= 0.01A

Voltmeter reading = Pd across R\(_2\)

= 0.01 x 400 = 4v

 

(c)

\(\frac{1}{R_p} = \frac{1}{2} + \frac{1}{2} = \frac{1 + 1}{2} = \frac{2}{2}\)

Note: This catadation is for the two resistors in parallel.

R\(_T = 2 + 1 + 2 = 5\Omega\)


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WAEC May/June 2024 - Practice for Objective & Theory - From 1988 till date, download app now - 99995