A capacitor charging circuit consists of a battery, an uncharged 20 μF capacitor, and a 5.0 kΩ resistor. At t = 0 s the switch is closed; 0.15 s later, the current is 0.62 mA.

What is the battery's emf?

emf = 0.437 V

Explanation:

given:  R= 6×10³Ω, t=0.15 s, C = 20  × 10 ⁻⁶ F, I= 0.62 × 10 ⁻³ A

Solution:

Using formula I = I₀ × /RC

⇒ I₀ = I RC /

I₀ = 0.62 × 10 ⁻³ A × 5×10³Ω × 20  × 10 ⁻⁶ F /

I₀ = 0.000062 /0.707

I₀ = 8.758 × 10⁻⁵ A

emf = I₀ × R

emf = 8.758 × 10⁻⁵ A × 5×10³Ω

emf = 0.437 V

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