Question

In an RC series circuit, ℰ = 17.0 V, R = 1.60 M, and C = 1.80 µF.

Answer 1

Answer:

(a) τ=2.88 s, (b) q₀= 30.6μC and (c) t=1.434s            

Explanation:

A RC circuit is an resistor(R)-capacitor(C) electric circuit.    

(a) In a resistor-capacitor circuit, the time constant (τ) can be calculated by:    

$\tau = RC$          

where R: is the resistence and C: the capacitance of the capacitor

$\tau = (1.60\cdot 10^{6} \cdot 1.80\cdot 10^{-6} = 2.88 s$  

 

(b) The maximum charge (q₀) is giving by:    

$q_{0} = \epsilon \cdot C$  

where ε: is the voltage across the capacitor

$q_{0} = 17.0 V \cdot 1.80 \cdot 10^{-6} F = 3.06 \cdot 10^{-5} C = 30.6 \mu C$                  

(c) The time (t) that take the charge (q) to build up to 12 μC can be calculated from the next equation:  

$q = q_{0}(1 - e(\frac{-t}{RC}))$  

$t = RC Ln( \frac{q_0}{q_0 - q}) = \tau Ln( \frac{q_0}{q_0 - q})$

$t = 2.88 \cdot Ln (\frac{30.6 \cdot 10^{-6}}{(30.6 \cdot 10^{-6} - 12.0 \cdot 10^{-6}})$

$t = 1. 434 s$  

Have a nice day!