Part G: Use Kirchoff's Law to solve for all the voltage drops and branch currents. R1 100 a R4 700 ww ww R2 225 Q Vs2 15V R5 125 0 Vs1 20V ww R3 Vs3 25 0 6V ww ww-
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- What would the potential difference in part A be if the wire were silver instead of copper, but all else was the same? ρsilver=1.47×10−8Ω⋅m�silver=1.47×10−8Ω⋅m.Figure 8₁ 9.0 V R₁ 22 2 R₂ = 1892 www 83 = 6.0 V HH V 1 of 1 Part A Determine the magnitudes of the currents through R₁ and R₂ in (Figure 1), assuming that each battery has an internal resistance r = 2.2 . Express your answers using two significant figures separated by commas. I₁, I2 = Submit Part B Submit V— ΑΣΦ Request Answer I₁ to the left; I2 to the left. I₁ to the left; I₂ to the right. I₁ to the right; I₂ to the left. I₁ to the right; I₂ to the right. Determine the directions of the currents through R₁ and R₂. Provide Feedback Correct Previous Answers 2 ? AThe capacitor with capacitance C₁ initially carries a charge Q (see figure below). While, the capacitor with capacitance C₂ is initially uncharged. The switch S is closed at time t = 0, and the charges begin to flow in the circuit. Find the charge flown through the wires as a function of time, q(t). QC₁ A) (0) (1 - e-(C₁+C₂)t/RC₁C₂) C₁+C₂ B) (02₂) (1 - e-(C₁+C₂)t/RC₁C₂) C₁+C₂ QC₁ C₂ C) (043) (1 — e¯(C₁+C₂)t/RC₂C₂) D) Q(1 + e-(₁+₂)t/RC1C₂) E) None of the above. +Q C₁ R C₂
- Switch S1 has been closed and S2 has open for a long time until the capacitors are fully charged. At t=0, S1 is opened and S2 is closed. If E=12V, C₁=4mF and C₂=8mF, R₁-3.22k0 and R₂-1.22 k0; what is the current passing through R₂ at t=0 in units of mA? Please express your answer using two decimal places. & R₁ www Answer: S₁ HH C₂ C₁. a b I S₂ R₂Problem 4, 1.10 Further Application Application of Kirchoffs law for the following mesh find curent I, L.I, I, using Ohm's Law Voltage (current) resistance with the conventionthat a curent loop away from + back %3D to as in mesh1 is consideredpositive voltage negative voltage as in mesh 2 as it loops from backto + 19 V 12 I 3Ω 32 291 20 I, 2. 14 V 10 40 -10 ww ww