Answer to Question #153554 in Electric Circuits for HARIS

Question

CONSIDER A CIRCUIT HAVING 30 VOLT SOURCE , SWITCH , RESISTANCE R , CAPACITOR 0.01 MICRO FARAD AND RESISTANCE 10 KILO OHM CONNECTED IN SERIES WHEN THE SWITCH IN THIS CIRCUIT IS CLOSED , THE VOLTAGE ACROSS THE CAPACITOR MUST RISE TO 6.59 VOLT IN 0.1 MS IDENTIFY THE VALUE OF R THAT SHOULD BE USED

Answer 1

Solution:

$v=v_o[1-e^{(\tfrac{-t}{\tau})}]$

6.59 = 30[1–$e^{(\tfrac{-0.0001}{\tau})}$]

1–$e^{(\tfrac{-0.0001}{\tau})}=0.2197$

$e^{(\tfrac{-0.0001}{\tau})}=0.7803$

$\dfrac{-0.0001}{\tau}=ln(0.7803)=-0.2481$

$\tau=\dfrac{0.0001}{0.2481}=0.0004031\;s$

RC = τ = 0.0004031

R = $\dfrac{0.0004031}{0.01e^{-6}}=40310\;\Omega$

added R = 40310 – 10000 = 30310 Ω

the voltage on a cap, charging

$v=v_o[1-e^{(\tfrac{-t}{\tau})}]$

v₀ is the battery voltage

v is the voltage after time t

R is resistance in ohms,

C is capacitance in farads

t is time in seconds

RC = τ = time constant