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3 years ago

14

A 7.8 µF capacitor is charged by a 9.00 V battery through a resistance R. The capacitor reaches a potential difference of 4.20 V

at a time 3.21 s after charging begins. Find R.

1 answer:

Vaselesa [24]3 years ago

3 0

Answer:

655128 ohm

Explanation:

C = Capacitance of the capacitor = 7.8 x 10⁻⁶ F

V₀ = Voltage of the battery = 9 Volts

V = Potential difference across the battery after time "t" = 4.20 Volts

t = time interval = 3.21 sec

T = Time constant

R = resistance

Potential difference across the battery after time "t" is given as

$V = V_{o} (1-e^{\frac{-t}{T}})$

$4.20 = 9 (1-e^{\frac{-3.21}{T}})$

T = 5.11 sec

Time constant is given as

T = RC

5.11 = (7.8 x 10⁻⁶) R

R = 655128 ohm

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