Answer:
The voltage is
![V_c = 9.92 \ V](https://img.qammunity.org/2021/formulas/physics/college/c5ph89jekeppgi0gzuu36qd5sh3ihr7bbp.png)
Step-by-step explanation:
From the question we are told that
The voltage of the battery is
![V_b = 24 \ V](https://img.qammunity.org/2021/formulas/physics/college/5wsz7nnq598kmmbaj34h4sd7o59wo7u8br.png)
The capacitance of the capacitor is
![C = 3.0 mF = 3.0 *10^(-3) \ F](https://img.qammunity.org/2021/formulas/physics/college/w71gdplluf7f0ngaophhud17pot65qq4zv.png)
The resistance of the resistor is
![R = 100\ \Omega](https://img.qammunity.org/2021/formulas/physics/college/4mlzeckmdis1vid9ohmetj15w3buvs40ag.png)
The time taken is
Generally the voltage of a charging charging capacitor after time t is mathematically represented as
![V_c = V_o (1 - e^{- (t)/(RC) })](https://img.qammunity.org/2021/formulas/physics/college/y2w0i2r13icev9gxsyco5byr6kbiixnoib.png)
Here
is the voltage of the capacitor when it is fully charged which in the case of this question is equivalent to the voltage of the battery so
![V_c = 24 (1 - e^{- (0.16)/(100 * 3.0 *10^(-1)) })](https://img.qammunity.org/2021/formulas/physics/college/oebqb8jq8tnh0gy2sa4z77h72pyo5arcyl.png)
![V_c = 9.92 \ V](https://img.qammunity.org/2021/formulas/physics/college/c5ph89jekeppgi0gzuu36qd5sh3ihr7bbp.png)