Question

An uncharged capacitor and a resistor are connected in series to a source of emf. If 12.0 v, c 19.0 μF, and R = 100 Ω, find the following. a) (3 points) the time constant of the circuit b) (2 points) the maximum charge on the capacitor c) (3 points) the charge on the capacitor at a time equal to one time constant after the battery is 7. connected. d) (2 points) the current at t = 2 s.

Answer 1

The given case is RC-charging.

Thus, the voltage as a function is given by

  $V_C=\varepsilon (1-e^{\frac{-t}{\tau}})$ ...... (1)

From given data,

i) resistance is $R=100 \Omega$

ii) capacitance is $C=19 \mu\textup{F}=19 \times 10^{-6}\textup{ F}$

iii) supplied voltage $\varepsilon =12\textup{ V}$

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a)

Time constant of an RC circuit is given by

$\tau=R \times C$

substitute the known values,

$\tau=100 \times 19 \times 10^{-6}$

$\tau=0.0019\textup{ sec}$

Thus, the time constant is $\boldsymbol{\tau=0.0019\textup{ sec}}$

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b)

Maximum charge on the capacitor is given by

$Q_{max}=CV_{max}$ ...... (2)

here $V_{max}$ is the maximum voltage of capacitor.

Maximum voltage on capacitor is obtained when time is 5times the time constant.

Therefore, from (1),

$V_{max}=\varepsilon (1-e^{\frac{-5 \tau}{\tau}})$

substiute the known values,

$V_{max}=12(1-e^{-5})$

$V_{max}=11.92\textup{ V}$

substitute this value in (2)

  $Q_{max}=19 \times 10^{-6} \times 11.92$

  $Q_{max}=0.2265 \times 10^{-3}\textup{ C}$

Thus, the maximum charge on the capacitor is $\boldsymbol{Q_{max}=0.2265 \times 10^{-3}\textup{ C}}$

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c)

Charge on the capacitor is given by

  $Q=CV_C$

From (1)

  $Q=C \varepsilon (1-e^{\frac{-t}{\tau}})$

Therefore, when $t=\tau$,

$Q=C \varepsilon (1-e^{\frac{-\tau}{\tau}})$

substitute all the known values,

$Q=19 \times 10^{-6} \times 12 \times (1-e^{-1})$

  $Q=0.144 \times 10^{-3}\textup{ C}$

Thus, the charge at time equal to one time constant is $\boldsymbol{Q=0.144 \times 10^{-3}\textup{ C}}$

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d)

Current is given by

$I=\frac{V_C}{R}$

From (1)

$I=\frac{\varepsilon (1-e^{\frac{-t}{\tau}})}{R}$

substitute all the known values,

$I=\frac{12 \times (1-e^{\frac{-2}{0.0019}})}{100}$

$I=0.12\textup{ A}$