In the figure, suppose the switch has been closed for a length of time sufficiently long for the capacitor to become fully charged. For this circuit, R1 = 12.0 kΩ, R2 = 15.0 kΩ, R3 = 3.000 kΩ, C = 10.0 μF , and emf = 9.00 V. Find (d) the potential differance across R2. (e) the charge on the capacitor. R1CR2ɛmfR3

Given The resistance on R1 is R1= 12 kΩ=12×10-3 Ω. The resistance on R2 is R2= 15 kΩ=15×10-3 Ω.…

Answered: In the figure, suppose the switch has…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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In the figure, suppose the switch has been closed for a length of time sufficiently long for the capacitor to become fully charged. For this circuit, R₁ = 12.0 kΩ, R₂ = 15.0 kΩ, R₃ = 3.000 kΩ, C = 10.0 μF , and emf = 9.00 V. Find (d) the potential differance across R₂. (e) the charge on the capacitor.

Expert Solution

Part (d)

Given

The resistance on R₁ is $R_{1} = 12 kΩ = 12 \times 10^{- 3} Ω$ .
The resistance on R₂ is $R_{2} = 15 kΩ = 15 \times 10^{- 3} Ω$ .
The resistance on R₃ is $R_{3} = 3 kΩ = 3 \times 10^{- 3} Ω$ .
The capacitance across a capacitor is $C = 10.0 μF = 10 \times 10^{- 6} F$ .
The emf of the circuit is $ε = 9 V$ .

When the switch has been closed then the capacitor to become fully charged due to which it acts as a short circuit and hence no current passes through the R₃. Therefore current only flows through R₁ and R₂.

By the voltage division rule, the potential difference (voltage ) across R₂is calculated as,

$\begin{array}{rcl} V_{2} & = & \frac{ε \times R_{2}}{R_{1} + R_{2}} \\ = & \frac{9 V \times 15 \times 10^{- 3} Ω}{12 \times 10^{- 3} Ω + 15 \times 10^{- 3} Ω} \\ = & 5 V \end{array}$

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