Answer:
SPST-2
Explanation:
The terms post and pull are also used to describe the contact variations of the switch. The number of "poles" is the number of electrically separated switches that are controlled by a single physical actuator.
For example,
The "SPST" switch is single-pole, single-throw, a simple on / off switch: the two terminals are connected or disconnected from each other.
The "SPDT" switch is single pole, double throw, a simple interrupt changeover switch before making: C (COM, Common) is connected to L1 (NC, normally closed) or L2 (NO, normally open).
"DPST" is bipolar, single shot, equivalent to two SPST switches controlled by one circuit.
"DPDT" is bipolar, double throw, equivalent to two SPDT switches controlled by two circuits.
"3PDT" is 3 pole, double throw, equivalent to 3 SP and DT switches controlled by two circuits.
Answer:
B. The circuit would have less resistance.
Explanation:
When the lightbulb is replaced in the circuit by a wire, the circuit will have less resistance.
Resistance is the the opposition to the flow of current within a circuit. Both a light bulb and wire are two examples of resistors in a circuit.
- A wire has a low value of resistance compared to the bulb.
- This is why the overall resistance will reduce when a light bulb is replaced with a wire.
The standard cell potential helps to determine the oxidative and reductive strength of species. All the species that lie below the Standard Hydrogen Electrode in the standard cell potential table are stong oxidizing agents and are highly electronegative while those that are above the Standard Hydrogen Electrode are either weak oxidants or reducing agents
Basically, it sends out a signal that is the same frequency as your phone. This signal is powerful enough to outmuscle your phone's signal.
Answer:
the thermal energy generated in the loop = 
Explanation:
Given that;
The length of the copper wire L = 0.614 m
Radius of the loop r = 
r = 
r = 0.0977 m
However , the area of the loop is :
Change in the magnetic field is 
Then the induced emf e = 
e = 
e = 2.74 × 10⁻³ V
resistivity of the copper wire 
Ω m
diameter of the wire = 1.08 mm
radius of the wire = 0.54 mm = 0.54 × 10⁻³ m
Thus, the resistance of the wire R = 
R = 
R = 1.13× 10⁻² Ω
Finally, the thermal energy generated in the loop (i.e the power) = 
= 
=
