Amali 1 Kapasitor.docx
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AMALI 1 : KAPASITOR Pengiraan : R= 47.1 Ω Kapasitor 1 C =27.11 μF 𝜏=RC =( 47.1 )( 27.11x10 –6 ) =1.277x 10–3 s Q= CV = ( 27.11 x 10–6 ) (3) = 8.133 x 10 –5 C Kapasitor 2 C=47 μF 𝜏=RC =( 47.1 )( 47x 10 –6 ) =2.214 x 10 –3 s Q= CV = ( 47 x 10–6 ) (3) = 1.41 x 10 –4 C Kapasitor 3 C= 50 μF 𝜏=RC =( 47.1)( 50 x 10 –6 ) =2.355 x 10 –3 s Q= CV = (50 x 10–6 ) (3) = 1.5 x 10 –4 C
Charging ΔV (t) = 𝜀 ( 1 − 𝑒 –1/𝜏 ) = 3 ( 1 – e –1/ 1.277 x 10-3 ) =3 V
ΔV (t) = 𝜀 ( 1 − 𝑒 –1/𝜏 ) = 3 ( 1 – e –1/ 2,214 x 10-3 ) =3 V
ΔV (t) = 𝜀 ( 1 − 𝑒 –1/𝜏 ) = 3 ( 1 – e –2.355 x 10-3 ) =3 V
Discharging ΔV (t) = ΔV0 𝑒 –1/𝜏 = 3 (𝑒 –1/1.277 x 10-3 ) =0 V
ΔV (t) = ΔV0 𝑒 –1/𝜏 = 3 (𝑒 –1/2.214 x 10-3 ) =0 V
ΔV (t) = ΔV0 𝑒 –1/𝜏 = 3 (𝑒 –1/2.355 x 10-3 ) =0 V
Charging ΔV (t) = 𝜀 ( 1 − 𝑒 –1/𝜏 ) = 3 ( 1 – e –1/ 1.277 x 10-3 ) =3 V
ΔV (t) = 𝜀 ( 1 − 𝑒 –1/𝜏 ) = 3 ( 1 – e –1/ 2,214 x 10-3 ) =3 V
ΔV (t) = 𝜀 ( 1 − 𝑒 –1/𝜏 ) = 3 ( 1 – e –2.355 x 10-3 ) =3 V
Discharging ΔV (t) = ΔV0 𝑒 –1/𝜏 = 3 (𝑒 –1/1.277 x 10-3 ) =0 V
ΔV (t) = ΔV0 𝑒 –1/𝜏 = 3 (𝑒 –1/2.214 x 10-3 ) =0 V
ΔV (t) = ΔV0 𝑒 –1/𝜏 = 3 (𝑒 –1/2.355 x 10-3 ) =0 V
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