Question

Consider a circuit having 30V 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 V in 0.1 ms. Identify the value of R that should be used.

Answer 1

Answer:

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Answer 2

Given :

Vs = 30V

C = 0.01μF

R1 = 10kΩ

Vc = 6.59V

t = 0.1ms

To find :

Resistance R

Solution :

$V_c(t) = V_s (1 - e^{\frac{-t}{\tau } })$

$6.59 = 30(1-e^{\frac{-0.1*10^{-3} }{ \tau } } )$

τ = RC

τ = (R + 10 × 10³)(0.01 × 10⁻⁶)

τ = (R × 0.01 × 10⁻⁶ + 100 × 10⁻⁶)

$\frac{6.59}{30} = (1 - e^{\frac{0.1*10^{-3} }{R * 10^{-8}+10^{-4} } } )$

$0.78 = e^{-(\frac{0.1*10^{-3} }{R*10^{-8} +10^{-4} } )}$

$log(0.78) = - (\frac{0.1*10^{-3} }{R*10^{-8}+10^{-4} } )$

$-0.107 = - (\frac{0.1*10^{-3} }{R*10^{-8}+10^{-4} } )$

$(R*10^{-8})+ (10^{-4}) = \frac{0.1*10^{-3} }{0.107}$

$(R *10^{-8})=(\frac{0.1*10^{-3} }{0.107} )-10^{-4}$

$R = (\frac{0.1*10^{-3} }{0.107*10^{-8} } -\frac{10^{-4} }{10^{-8} } )$

R = 0.9345 × 10⁵ - 10⁴

R = 8.34 × 10⁴Ω

R = 83.4kΩ