Three bulbs A circuit is made of two 1.6 volt batteries and three light bulbs as
ID: 2039166 • Letter: T
Question
Three bulbs A circuit is made of two 1.6 volt batteries and three light bulbs as shown in the figure. When the switch is closed and the bulbs are glowing, bulb 1 has a resistance of 8 ohms, bulb 2 has a resistance of 45 ohms, bulb 3 has a resistance of 26 ohms, and the copper connecting wires have negligible resistance. You can also neglect the internal resistance of the batteries. Bulb 2 Bulb N M Bulb 3 (a) With the switch open, indicate the approximate surface charge on the circuit diagram.(Do this on paper. Your instructor may ask you to turn in this work.) Refer to your diagram to decide which of the following statements about the circuit (with the switch open) are true There is a large gradient of surface charge between locations M and L. There is no excess charge on the surface of the wire at location C. The surface charge on the wire at location B is positive. The electric fleld in the air between locations B and C is zero. The electric fleld in the filament of bulb 3 is zero. (b) With the switch open, find these potential differences: (c) After the switch is closed and the steady state is established, the currents through bulbs 1, 2, and 3 are 11, 12, and I3 respectively. Which of the following equations are correct loop or node equations for this steady state circuit? 1I I3 012 = 13 0-11"(8 ?)-12"(45 ?) + 13"(26 ?)-0 2 (452) I3 (26 12)0 +3.2V +-11"(8?) + 13"(8?)-0 (d) In the steady state (switch closed), which of these are correct? (f) Now find the unknown currents, to the nearest milliampere. (L.e. enter your answer to three decimal places.) (g) How many electrons leave the battery at location N every second? (i) What is the numerical value of the power delivered by the batteries? (j) The tungsten filament in the 45 ohm bulb is 11 mm long and has a cross-sectional area of 2 × 10-10 m2, what is the magnitude of the electric field inside this metal electrons/s filament? V/mExplanation / Answer
(f)
Parallel combination of bulbs 2 and 3 has a resistance = 16.47 Ohm (=R2*R3/R2+R3).
Total resistance = R1 + (R2 parallel R3) = 24.47 Ohm,
hence
I1 = 3.2 / 24.47 = 130 mA (answer)
Now voltage-divider law tells us that the voltage across (R2 parallel R3)= 3.2*16.47/24.7 = 2.153 Volt.
Current through each bulb is this voltage by corresponding resistance
I2 = 2.153/45 = 47 mA (answer)
I3 = 2.153/26 = 82 mA (answer)
As a sidelight (no pun intended), I1 = I2 + I3 to within desired accuracy (129 mA), as required.
(g) Current leaving batteries = i1 = 130 mA.
By definition, i1 = q*N where q is charge on an electron = 1.6*10-19 C and
N = number per sec = 8.13*1017 electrons per sec.
(i) P = IV = 0.130*3.2 (he says batteries - plural) = 0.416 Watts (answer)
(j) The length of the filament = 11*10-3 m, voltage across = 2.153 Volt as we found out using the voltage divider rule.
Field = V/L = 2.153/11*10-3 = 195.732 V/m.
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