Question

A 1.70-μF capacitor is charging through a 14.0-Ω resistor using a 10.0- V battery. What will be current when the capacitor has acquired 1/4 of its maximum charge?

Answer 1

Answer:

I = 0.536 A

Explanation:

The maximum charge of the capacitor is ten volts because after some time the capacitor will acquire the same potential than the battery (Vb), therefore, a quarter  of the maxim voltage (10V) is 10V/4 =2.5V.  

To solve the problem we need to apply Ohm's law  V = I*R when the capacitor  voltage (Vc) is at 2.5V, as you can see from the image of the circuit the resistance is between the capacitor and the battery, therefore, the potential difference across the resistance (Vr) is ten volts of the battery minus two point five volts of the capacitor -> Vr = Vb-Vc = 10V-2.5V = 7.5V, applying Ohm's law to the resistance Vr = I*R -> I = Vr/R = 7.5V/14Ω = 0.536 A.

Note: Because all the elements on the circuit are in series the current in all components is the same, please see the graph current vs voltage of the simulated circuit and notice that the current calculated matches the simulation.

Answer 2

Answer:

The current is 0.951 A

Solution:

As per the question:

Capacitance of the capacitor, C = $1.70\mu F$

Resistance of the circuit, R = 14.0$\Omega$

Voltage of the battery, V = 10.0 V

Now, when the charge become one-fourth of its initial value, the amount of current flowing is given by:

$q' = q(1 - e^{-\frac{t}{\tau}})$          (1)

where

q' = $\frac{q}{4}$

$\tau = RC$

$q' = q(1 - e^{-\frac{t}{RC}})$

$\frac{q}{4} = q(1 - e^{-\frac{t}{RC}})$

$0.25 = (1 - e^{-\frac{t}{RC}})$

$0.75 = e^{-\frac{t}{RC}$

$-\frac{t}{RC} = ln(0.75)$

$-\frac{t}{14\times 1.70\times 10^{- 6}} = - 0.2876$

$t = 6.844\times 10^{- 6} s$

Now,

The current is given by:

$I' = Ie^{-\frac{t}{RC}}$                  (2)

Also, $I = \frac{V}{R} = \frac{10}{14} = 0.714 A$

Now, using the value of I = 0.714 A

$I' = 0.714e^{-\frac{6.844\times 10^{- 6}}{14\times 1.70\times 10^{- 6}}} = 0.951 A$