Three bulbs A circuit is made of two 1.6 volt batteries and three light bulbs as

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 6 ohms, bulb 2 has a resistance of 38 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 Bulb 1 0 Bulb 3 Povernel differnid Sm f volt.of Laten (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)

Explanation / Answer

  a)

The two statements which you've marked true, are correct. This does assume that the entire system is "floating" relative to ground, so that no excess charge can go elsewhere.

1.There is no excess charge on the surface of the wire at location C is FALSE statement

Unless you ground the negative terminal, the negative half of your circuit will be just as negative, as the positive half is positive. Usually, we ground one terminal, so that its charge excess is distributed to the rest of Earth, and so that it is touch-safe.

2.The surface charge on the wire at location B is positive is TRUE statement .This does

assume that the entire system is floating relative to ground. so that no excess can go

elsewhere.

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

4.The electric field in the air between locations B and C is zero is FALSE statement
This will be where the electric field is. B will be at the batteries' total positive voltage, and C will be at the batteries' negative voltage. This differential in voltage is what will become the electric field across the air gap of the switch.

5.There is a large gradient of surface charge between locations M and L is also false statement.

Once electrostatic equilibrium is achieved, all conductors distribute their surface charge to get the surface charges as far from one another as possible. It doesn't matter if they are low resistivity copper, or high resistivity tungsten

g) answer

how many electrons leave the battery at location N every second

the current flowing at location N, is I1. It is flowing in to the battery, thus electrons are exiting the battery. We need to translate from amperes per second, into elementary charge units per second, since the charge of 1 electron is 1 elementary charge unit.

1 elementary charge unit is 1.602x10-19 Coulombs.

And the Ampere is 1 Coulomb per second. Believe it or not, the ampere came first in the original definitions.

Divide by the elementary charge unit, in Coulombs
0.147 amps = 9.17x1017 electron/sec
Thus, 9.17x1017 electrons exit the negative battery terminal each second.

j)

Electric field is voltage divided length, assuming that it is uniform. In a wire such as this, we might as well assume it is uniform. The cross sectional area doesn't really matter, all that does is assist us in confirming that it is 38 ohms.

E = V/L

V = I2xR2
E = I2xR2/L

Already calculated I2=0.046A in question f

Given R2=38 ,L=12mm=0.012m
E = (0.046)x(45/0.012 )

=172.5 N/C
|E| = 172.5 V/m

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