A circuit is constructed with five resistors and one real battery as shown above

Question

A circuit is constructed with five resistors and one real battery as shown above right. We model. The real battery as an ideal emf V = 12 V in series with an internal resistance r as shown above left. The values for the resistors are: R1 = R3 = 27 , R4 = R5 = 116 and R2 = 134 . The measured voltage across the terminals of the batery is Vbattery = 11.72 V.

1)What is I1, the current that flows through the resistor R1?

mA

2)What is r, the internal resistance of the battery?

3)What is I3, the current through resistor R3?

mA

4)What is P2, the power dissipated in resistor R2?

W

5)What is V2, the magnitude of the voltage across the resistor R2?V

6)

Resistor R2 is now shorted out as shown. How does the magnitude of the voltage across the battery change? [ PICK THE CORRECT ANSWER ]

1) Vbattery decreases

2)Vbattery increases

3)Vbattery remains the same

Explanation / Answer

given that

R1 = R3 = 27 ohm , R4 = R5= 116 ohm

R2 = 134 ohm

Vb = 11.72 V

E = 12 V

from the figure

R3 ,R4 , R5 are in series

so Rs = 27+116+116 = 259 ohm

R2 || Rs

so Rp = 134*259 / (134+259) = 88.31 ohm

R1 and Rp are in series

so Rt = 27+88.31 = 115.31 ohm

current i = V/Rt

i = 11.72/115.31 = 0.10 A

(a)

i1 = i = 0.10 A

(b)

r = (E-V) / i

r = (12-11.72) / 0.10

r = 2.8 ohm

(c)

i3 = V - (i*R1) / Rs

i3 = 11.72 - (0.10*27) / 259

i3 = 0.034 A

(e)

V2 = V - R1* i = 11.72 - ( 27*0.10)

V2 = 9.02 V

(d)

P2 = power dissipated in resistor R2 = V2^2 / R2

P2 = (9.02)^2 / 134 = 0.60 W

(f)

If R2 is shorted out , then just R1 remains with r in series

Vb' = E*R1 / (R1+r)

Vb' = 12*27 / (27+2.8) = 10.87 V

so we can see that battery voltage is decreased

answer(1) is correct

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