Determine potential difference across capacitor at t=π16sec. {Emf of AC source is in volts and ‘t’ is in seconds]

Z¯=R+XL¯Xc¯XL¯+Xc¯=100+20j×25(−j∧)20j−25j=100(1+j)i¯=ε¯Z¯ ∴v¯C=i¯(Z¯−R)=100i(4t−π4)100(1+j).100j=102ei(4t−π4)(1+j)=52ej4tAt t=π16S;vC=5V

Book Online Demo

Check Your IQ

Try Test

Courses

Determine potential difference across capacitor at $t = \frac{π}{16} \sec$ . {Emf of AC source is in volts and ‘t’ is in seconds]

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

answer is 5.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

$\bar{Z} = R + \frac{\bar{X_{L}} \bar{X_{c}}}{\bar{X_{L}} + \bar{X_{c}}}$

$= 100 + \frac{20 j \times 25 (- \overset{\land}{j})}{20 j - 25 j} = 100 (1 + j)$

$\bar{i} = \frac{\bar{ε}}{\bar{Z}} ∴ {\bar{v}}_{C} = \bar{i} (\bar{Z} - R) = \frac{100^{i (4 t - \frac{π}{4})}}{100 (1 + j)} .100 j$

$= \frac{10}{2} e^{i (4 t - \frac{π}{4})} (1 + j) = 5 \sqrt{2} e^{j 4 t}$

At $t = \frac{π}{16} S; v_{C} = 5 V$

Watch 3-min video & get full concept clarity