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Consider a resistor (R=1000 kΩ) and a capacitor (C=1μF) connected in series. This configuration is connected in series to a battery with an emf of 10 V.1) Find the maximum voltage in [V] across the capacitor.2) Find the maximum charge on the capacitor in [µC].3) How long will it take until the capacitor has a potential difference of 5 V 4) After the capacitor is charged to its maximum value, the capacitor and resistor are disconnected from the battery and from each other. Then they are connected in parallel, so that the capacitor discharges through the

Consider a resistor R1000 kΩ and a capacitor C1μF connected in series This configuration is connected in series to a battery with an emf of 10 V1 Find the maxim class=

Respuesta :

Given data:

The resistance is R=1000 kΩ.

The capacitance of capacitor is C=1 μF.

The emf of battery is V₀=10 V.

Part (1)

The maximum voltage in the capacitor reached would be equal to the voltage of battery, i.e., V=10 V.

Thus, the maximum voltage of battery is V=10 V.

Part (2)

The maximum charge on the capacitor can be calculated as,

[tex]\begin{gathered} Q=VC \\ Q=(10\text{ V)(1}\times10^{-6}\text{ F)} \\ Q=10^{-5}\text{ C} \end{gathered}[/tex]

Thus, the maximum charge on capacitor is 10⁻⁵.

Part (3)

The time taken to reach the voltage 5 V will be,

[tex]\begin{gathered} V=V_0(1-e^{-\frac{t}{RC}}) \\ 5=10(1-e^{-\frac{t}{(1000)(10^{^{-6}})}}) \\ 0.5=e^{-\frac{t}{(1000)(10^{-6}^{}^{})}} \\ t=0.693\times10^{-3}\text{ s} \end{gathered}[/tex]

Thus, the time taken is 0.693x10⁻³ s.

Part (4)

The time taken can be calculated as,

[tex]\begin{gathered} V=V_0(1-e^{-\frac{t}{RC}}) \\ 2.5=10(1-e^{-\frac{t}{(1000)(10^{^{-6}})}}) \\ 0.75=e^{-\frac{t}{(1000)(10^{-6}^{}^{})}} \\ t=2.11\times10^{-3}\text{ s} \end{gathered}[/tex]

Thus, the time taken is 2.11x10⁻³ s.