(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.
(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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Correct Answer: Option n
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\(_T = 2 + 1 + 2 = 5\Omega\)
(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\(_T = 2 + 1 + 2 = 5\Omega\)