Three bulbs A circuit is made of two 1.4 volt batteries and three light bulbs as

Question

Three bulbs A circuit is made of two 1.4 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 13 ohms, bulb 2 has a resistance of 39 ohms, bulb 3 has a resistance of 27 ohms, and the copper connecting wires have negligible resistance. You can also neglect the internal resistance of the batteries Bulb 2 Balb 1 N M Bulh (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 B The surface charge on the wire at location B is positive. O There is no excess charge on the surface of the wire at location C There is a large gradient of surface charge between locations M and L The electric field in the air between locations B and Cis zero. O The electric field 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 I I2, and I3 respectively. Which of the following equations are correct loop or node equations for this steady state circuit? -11*(13 ?)-?*(390) + 13*(27 ?)-0 11 12 +13 -12 (39 13(27o)0 O +2.8V 13 +12 (39)0 (d) In the steady state (switch closed), which of these are correct? (F) Now find the unknown currents, to the nearest milliampere. (I.e. enter your answer to three decimal places.) (g) How many electrons leave the battery at location W every second? (i) What is the numerical value of the power delivered by the batteries? i) The tungsten filament in the 39 ohm bulb is 5 mm long and has a cross-sectional area of 2 x 1010 m2. What is the magnitude of the electric field inside this metal filament? electrons/s Vim

Explanation / Answer

(a) From the given option, correct options are :

The electric field in the filament of bulb 3 is zero.

The surface charge on the wire at location B is positive.

(b) VB - VC = 2*1.4 = 2.8 V

  VD - VK = 0 V

(c) By looking at the diagram and using junction rule, we have

I1 = I2 + I3  

and using kirchoff's loop rule (whole loop), we get

2.8 - 13*I1 - 39*I2 = 0

Now the third loop equation will contain only bulb 2 and 3, we get

-39*I2 + 27*I3 = 0

SO, THIRD, FOURTH AND FIFTH OPTIONS ARE CORRECT.

(d) Now, when switch is closed and we have a steady state, we need to consider only those elements which lie between 2 points

then VC - VF = 39*I2  

VC - VF = 27*I3

VL - VA = -2.8 + 13*I1

SO OPTIONS FIRST , SECOND AND LAST ARE CORRECT.

(e) This one is very easy, Just find the equivalent resiatance first.

combine both parallel resistances, we get R = 15.9545

THEN combine this in series with R1 , WE GET Requivalent = 28.9545

so, I1 = V /  Requivalent = 2.8 / 28.9545 = 0.0967 amps

To find I2 and I3 , We will find voltage across them, V = IR = 0.0967*15.9545 = 1.543 volts,

so, I2 = 1.543 / 39 = 0.0396 amps

I3 = 1.543 / 27 = 0.05715 amps

(g) Number of electron = 0.0967 / 1.602e-19

= 6.036e17 electrons/second

(i) P = VI

P = 0.0967*2.8 = 0.2708 watt

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