What is the power dissipated in R
Effect of adding R5 = 3 * R2 to the design on the current in R2, IR2 :
see multiple choices on the picture they are the same for the second question
Reminder: R4 starnds for a potertiameter; for the current problem it is equivalent with two resistors R4a and R4b which meet the condition R4=R4a+R4b. For example, when the polentionteber is turned up halfway, R4a=R4b=(1)/(2)*R4.
R4 at 10% mears that R4a (the higher part of the potertiometer) is 10% of the tatal
value of the potertiameter R4, e.g. R4a=0.1**R4 and R4b=0.9**R4
What is the power
dissipated in RL?
Effect of adding R5 =
3^(+)R2 to the design
The current in R2 will triple:
Choose.-
Power dissipated in RL is 0.42mW
The cument in R2 will triple
The effect on the current in R2 is not listed among aptians
Power dissipated in RL is 1.42mW
The current in R2 will reduce to a third
Power dissipated in RL is 2.42mW
an the current in R2,
IR2 :
V1 R1 R2 R4 R4 % R4a R4b RL
1.00V 1.00k 2.00k 4.00k 40% 1.60kQ 2.40k 5.78k
R
Reminder: R4 stands for a potentiometer; for the current problem it is equivalent with two resistors R4a and R4b which meet the condition R4 = R4a + R4b. For example, when the potentiometer is turned up halfway, R4a = R4b 1/2 * R4 R4 at 10% means that R4a (the higher part of the potentiometer) is 10% of the total value of the potentiometer R4, e.g. R4a = 0.1 * R4 and R4b = 0.9 * R4
What is the power dissipated in RL?
The current in R2 will triple
Choose.. Power dissipated in RL is 0.42mW The current in R2 will triple The effect on the current in R2 is not listed among options Power dissipated in RL is 1.42mW The current in R2 will reduce to a third Power dissipated in RL is 2.42mW
Effect of adding RS = 3 * R2 to the design on the current in R2, IR2 :