Question

!PHYSICS! Anyone willing to share their modest mental qualities you are welcomeXD I will give owl bucks

Answer 1

No.

Answer 2

1. Since each light bulb carries a resistance of 120 Ω, the total resistance of series circuit is calculated as follows,

Rs=120Ω+120Ω+120Ω+120Ω+120Ω=600Ω

The total resistance in this series circuit is therefore 6.0*10^2.

On the other hand, if the light bulbs were placed in parallel manner, then the total resistance will be,

1/R=1/120+1/120+1/120+1/120+1/120

Isolating for R

1=R(1/120+1/120+1/120+1/120+1/120)
1=0.04166R

R=24.000

Therefore the total resistance in parallel circuit is 24Ω

2. From the diagram, junctions of the circuit are
a. H and B
b. Loops of this circuit consist of the loop ABCH and BCHDFG
1.

The electrical potential difference for the first diagram is calculated by reasoning with the fact that in series circuits voltages will be distributed in such ways that add up to the voltage at the battery depending on the resistance on each load.

Therefore,

Delta V2=60V-2.5V-1.1=56.4V

Therefore the voltage at Delta V2 is 56.4V


Since the voltage can be added to find the voltage at battery,

5.4V+7.5V+4.2V=17.1V

Therefore the electrical potential difference at battery is 17.1V

Answer 3

My answers owe thanks to the following questions

Answer 4

Six 120 Ω light bulbs are connected in series. What is the total resistance of this circuit? What would be the total resistance if they were connected in parallel?

Answer 5

that is a disappointing result @abb0t

Answer 6

Good try though Mr. Boo

Answer 7

XDXDXD

Answer 8

Perhaps @pooja195 can help since she seems to be so obsequious about medals.

Answer 9

@abb0t have nothing to say? DONT SAY IT.
I havent taken physics yet so i cant help sorry.

Answer 10

OMG, no diagrams?

Answer 11

@pooja195 He was jokingXD

Answer 12

50 tags http://prntscr.com/84lxqy NOT a joke hes being annoying