In the circuit at left, light bulbs A and B are identical, and each can be consi

Question

In the circuit at left, light bulbs A and B are identical, and each can be considered as a resistor with resistance R. When the switch is closed- (a) Does bulb B brighten, dim, or stay the same? ____ (b) If it dims or brightens, by what factor: ____ (remember, bulb brightness is proportional to power dissipated) (c) Does bulb A brighten, dim, or stay the same? ___ (d) With the switch closed there are three paths for current. Label the current in the 3V source as I_1, positive downward. Label the current in the lower lightbulb (B) as I_2, positive downward. Label the current in the other four components as I_3, positive in the sense shown in the diagram. With currents thus defined, write down Kirchoff's voltage law for the two principal loops of the circuit (containing the 3V supply and bulb B, and that containing the three 6 V sources and bulbs A and B). You may go either way around these loops but your signs must be consistent: (e) If R = 1.5 Ohm, solve for all the currents (you may need another equation):

Explanation / Answer

Q4.

when switch is open, current through the two bulbs=voltage/resistance

=(6+6+6)/(R+R)

=18/(2*R)

=9/R A

voltage across bulb B=current*resistance=(9/R)*R=9volts

when switch is closed, voltage across bulb B=3 volts

as power =voltage ^2/resistance

if voltage decreases, power decreases and the bulb dims.

part a:

bulb B dims

part b:

factor by which it dims=inital power/final power

=initial voltage^2/final voltage^2

=9^2/3^2=9

so it dims by a factor of 9.

part c:

voltage across A and B=18 volt

voltage across B=3 volts

then voltage across A=18-3=15 volts

as voltage across A increases, it brightens.

part D:

kirchoff' voltage law for left loop:

3-I2*R=0...(1)

kirchoff's voltage law for right loop:

18+I3*R-I2*R=0...(2)

part E:

from equation 1,

I2=3/R=3/1.5=2A

from equation 2:

18+I3*1.5-2*1.5=0

==>I3=(3-18)/1.5=-10 A

using KCL at the junction of I1 , I2 and I3:

I1+I2+I3=0

==>I1=-(I2+I3)=-(2-10)=8 A

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