question A battery with epsilon = 5.00 V and no internal resistance supplies cur
ID: 2276044 • Letter: Q
Question
question
A battery with epsilon = 5.00 V and no internal resistance supplies current to the circuit shown in the figure below. When the double-throw switch S is open as shown in the figure, the current in the battery is 1.00 mA. When the switch is closed in position a, the current in the battery is 1.24 mA. When the switch is closed in position b, the current in the battery is 1.82 mA. Find the resistance R1. Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. kohm Find the resistance R2. Find resistance R3.Explanation / Answer
Ohm's Law : V=IR
Case1 - switch is open
Total Resistance = R1+R2+R3
V=I(R1+R2+R3)
5 = (1*10^-3)(R1+R2+R3).........................(1)
Case2 - switch in position a
Total Resistance = R1+0.5R2+R3
V = I(R1+0.5R2+R3)
5 = (1.24*10^-3)(R1+0.5R2+R3).................(2)
Case 3 -switch in position b
Total Resistance = R1+R2
V = I(R1+R2)
5 = (1.82*10^-3)(R1+R2)...........................(3)
Solve (1),(2) and (3) we get
R1 = 0.81 kohms
R2 = 1.94 kohms
R3 = 2.25 kohms
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.