Need help capacitor discharge question

Thread Starter

wertyu

Joined Jul 13, 2020
10
I got it wrong twice and i made a silly error now im on last chance to pass .

can someone check my answer and given advice for me , it would help a lot .

thanks

capacitor discharge

10v supply - total resistance in circuit 10.3 Kohms - 100microfarad capacitor .

need to find Volt and Current at 1 sec and 2 secs .



below is my work

Voltage = 10 x ( e -t/tc)

1 sec = 10 x 0.379 = 3.79v

2 sec = 10 x 0.14 = 1.4v

Current Io x (e-t/tc)

1 sec 0.00097 x 0.379 = 0.000367a

2 sec 0.00097 x 0.1437 = 0.000143a
 

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Alec_t

Joined Sep 17, 2013
11,417
Welcome to AAC!
Re-check your equations for voltage and current. They are incorrect.
Show each stage of your working.
 

Thread Starter

wertyu

Joined Jul 13, 2020
10
ok
time constant = 10300 ohms x 0.0001 microfarad = 1.03 secs

Voltage

1 sec 10V x e to the power of minus (1 / 1.03 ) 0.97 = 0.379 10 x 0.379 = 3.79V

2 sec 10V x e to the power of minus ( 2 / 1.03 ) 1.94 - = 0.14 10 x 0.14 = 1.4 V

Current

10 / 10300 = 0.00097 amp

0.00097 x e to the power of minus 1 / 1.03 0.97 = 0.379 = 0.00097 x 0.379 = 0.000367a 0.367ma

0.00097 x e to the power of minus 2 / 1.03 1.94 = 0.14 = 0.00097 x 0.14 = 0.00013 0.1ma
 

Alec_t

Joined Sep 17, 2013
11,417
0.0001 microfarad
No. Carelessness with units will lose you marks.

You are still not showing all your working, which can lead to confusion. This line, for example:
"1 sec 10V x e to the power of minus (1 / 1.03 ) 0.97 = 0.379 10 x 0.379 = 3.79V"
would read more clearly as:
"At 1 sec, capacitor voltage = 10 x (e^-1/RC) = 10 x (e^-1/1.03) = 10 x e^-0.97 = 10 x 0.379 = 3.79V"
As the voltage at 1 sec is expressed to two decimal places your tutor may expect the voltage at 2 sec to be expressed likewise, i.e. as 1.44V.
 
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