Question

Help determining capitance and uncertainty ASAP

С, (10° F) | Cm (10° F) 2.8 2.58 СА Св 4.4 4.83

Answer 1

Solun : Given that - (z (10-2HF) cm (10 8 HF) CA 2.58 CB 4.4 4.83 For Capacitor A - A A co = 3+ + <- 2.8 + 2.58 +2.58) X10 = HE 2 A CA 11 2.69 X 10 3 SMF A bx | Cå - Ch1 = cm - col = 288103/HF A А 2.69x765 0.118103 НЕ for capacitor A x + 6 = (2.69 ± 0.11)X10²4 Similarly for capacitor & Cgt Can (4.4 +4.63) X10 u B Ca 2 2 B 4.698 10 -3 HF 11 4.62X10-3MF 6 = 16 - ;) = 14.62 - 44 / X15² MA B - bx 0.22 x 10 3MF

Now for B- C z 6 = (4.62 ± 0.22) x103 pf between wheat stone and Dmm Comparison For A - ACA = 1 cm - C3) = 26th B for B- AC₂ = 1 cm-C) = 26 * We Can allo calculate the rest Capacitance Capacitor C & D Wheatstone Vahe by using bridge relationship - CA =k CB co where k is some Conftant. CA 2.8 X10 3 HF 4.4810 3MF 0.64C, ²0.64 = CB ca then - if - s.0 x103 HF than Cc 3. 188103 Mf