In the circuit below, all four resistors are identical ( R 1 = R 2 = R 3 = R 4 =

Question

In the circuit below, all four resistors are identical (R1 = R2 = R3 = R4 = R) and the battery has a voltage of 5.01 V.

In the circuit below, all four resistors are identical (R1 = R2 = R3 = R4 = R) and the battery has a voltage of 5.01 V. When the switch is placed in position 1, the measured current in the battery is 1.39 mA. What is the value of each resistor? When the switch is placed in position 2, what is the current flowing out of the battery? When the switch is open (neither in position 1 or position 2), what is the current flowing out of the battery?

Explanation / Answer

a)

i=v/r

where r=R1+R4=2R

R=v/2i=3.604 kOhm

b)

r=R1+R2+R3//R4=2R+0.5R=2.5*3.604=9.010 kOhm

i=0.556 mA

c)

r=R1+R2+R4=3R=10.182kOhm

I=0.4663 mA

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