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 5.15 V.

In the circuit below, all four resistors are identical (R1 = R2 = R3 = R4 = R) and the battery has a voltage of 5.15 V. When the switch is placed in position 1, the measured current in the battery is 1.34 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)when switch is placed at position 1 R1,R4 are in series and R2 is shorted

Req= (R+R)= 2R= (5.15/1.34/0.001)

so R= 1.921kilo ohms

(b) when switch is placed in position 2 we get

R3,R4 are in parallel and this combination is in series with series combination of R1,R2

so Req= (R*R/2R) + R+ R = 5*R/2 = 4.802kilo ohms

current= Voltage/Req = 5.15/4.802K= 1.072mA

(c)when switch is off R2,R1,R4 are in series

Req =R1+R2+R4 =3R=5.76Kohms

I=5.15/(5.76*10^3)

I=0.894 mA

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.