Three bulbs A circuit is made of two 1.5 volt batteries and three light bubs as

Question

Three bulbs A circuit is made of two 1.5 volt batteries and three light bubs as shown in the figure. When the switch is closed and the bulbs are glowing, bulb 1 has a resstance of 9 ohms, bub 2 has a resistance of 45 ohms, bulb 3 has a resistance of 32 ohms, and the copper connecting wires have negigible resistance. You can also neglect the intemal resistance of the battenies. Rub 2 BulbI Bulb 3 (o) With the switch open, indicate the approximate surface charge on the circuit dagram.(Do this on paper. Your instructor may ask you to tun in thhs work.) Refer to your diagram to decide which of the following statements about the circuit (with the switch open) are true: There is no excess charge on the surface of the wire at location C. The electric field in the ar between locations B and C is zero. The surface charge on the wire at location B is positive. There is a large gradient of surface charge between locations M and L The electric field in the filament of bulb 3 is zero. (b) With the switch open, find these potential differences

Explanation / Answer

f) I1 = 0.108 A

I2 = I1*R3/(R2 + R3)

= 0.108*32/(45 + 32)

= 0.0449 A

I3 = I1*R2/(R2 + R3)

= 0.108*45/(45 + 32)

= 0.0631 A

g)

we know, I = q/t

I = N*e/t

==> N/t = I/e

= 0.108/(1.6*10^-19)

= 6.75*10^17 electrons/second

i) P = V*I

= 2*1.5*0.108

= 0.324 W

j)
we know,

use, R = rho*L/A

==> rho = R*A/L

= 45*2*10^-10/(8*10^-3)

= 1.126*10^-6 ohm.m

E = J*rho

E --> elctric field

J --> current density

rho --> resistivity

so,

E = (I/A)*rho

= (0.0449/(2*10^-10))*1.126*10^-6

= 253 N/c

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