Flag Electromagnetic Induction> A capacitor of capacitance 10 µF is conne...
question mark

A capacitor of capacitance 10 μF is connected to a battery of emf 2 V. It is found that it takes 50 ms for the charge on the capacitor to become 12.6 μC. Find the resistance of the circuit.

Radhika Batra , 11 Years ago
Grade 11
anser 1 Answers
Kevin Nash
Sol. C = 10 μF = 10^–5 F, emf = 2 V t = 50 ms = 5 * 10^–2 s, q = Q(1 – e^–t/RC) Q = CV = 10^–5 * 2 q = 12.6 * 10^–6 F ⇒ 12.6 * 10–6 = 2 * 10^–5 (1 e^-5 10 2 /R 10 5 ) ⇒12.6 * 10^-6/2 * 10^-5 = 1 – e^-5*10^-2 /R*10^-5 ⇒1 – 0.63 = -^-5*10^3 /R ⇒ -5000/R = In 0.37 ⇒ R = 5000/0.9942 = 5028 Ω = 5.028 * 10^3Ω = 5 KΩ.
Last Activity: 11 Years ago
star
LIVE ONLINE CLASSES

Prepraring for the competition made easy just by live online class.

tv

Full Live Access

material

Study Material

removal

Live Doubts Solving

assignment

Daily Class Assignments