user50642: I have an Arduino mega that is being used to control relays and ultrasonic sensors.  The relays that I am using is a 4 DPDT Signal Relay Module Model: D-256 Series and the other is a 4 channel Single pole Relay for Arduino. I am only using 3 channels of each relay. At any given time I only output to 2 channels from each relay at a time. I tested the relay to ensure they worked with the Arduino and they ran fine. 
Once I connected the Ultrasonic sensors to the board as well, the board would not power and the voltage regulator got extremely hot. I would assume the total current draw was too high from the outputs and VCC of the Arduino. 
I think that if I take my 9V battery and run it through a voltage regulator 7805 then I can run the VCC power of the relays with the battery instead of the VCC from the Arduino. This would result in me only running a VCC to the 4 ultrasonic sensors, the triggers for the ultrasonics from the Arduino, and the output signals fora maximum of 2 channels per relay board resulting in 4 relays. Most of the time it is only 2 relays. 
Would I be able to power the relay boards VCC with the 9V running through a 5v regulator instead of the arduino so that I can reduce the current running through the Arduino? 

user313: It sounds like a power-to-ground short circuit, either a fault in the Ultrasonic sensor or in the connection to the Arduino.
Check your connections, and if you have an Ohm-meter, check the resistance between the sensor's power and ground pins.
Current consumption is a function of the device being powered, not of the power-supply, unless the supply lacks the capacity to provide as much current as the device needs.
Note that 9v batteries (the small brick aka a "transistor battery") is not well suited to supplying much current and probably will not power your Mega, or not for long, anyway. They are much better suited for long-term powering of very low current devices, such as a smoke detector.

user42474: What could be going on here apart from what @JRobert has suggested, is, LM7805 could be generating a lot of heat loss.
4 X Ultrasonic sensors (SR HC-04) would draw 60mA.
2 X Relays (SRD-05VDC-SL-C) would draw 71.4 X 2 i.e 142.8mA.(Depends on from where you signal it)
The Arduino mega would draw anywhere between 200 - 500 mA.(even more)
The amount loss you will have is (9-5)V * (202.8 + 250)mA 
i.e. around 2 Watts. This is a lot of heat. I'm not even sure the LM7805 would be able to handle it.
Solution: use a switching regulator like the one mentioned here: https://electronics.stackexchange.com/questions/21131/efficient-low-power-regulation-i-e-9-5-volts

50642: <p>I have an Arduino mega that is being used to control relays and ultrasonic sensors.  The relays that I am using is a 4 DPDT Signal Relay Module Model: D-256 Series and the other is a 4 channel Single pole Relay for Arduino. I am only using 3 channels of each relay. At any given time I only output to 2 channels from each relay at a time. I tested the relay to ensure they worked with the Arduino and they ran fine. </p>

<p>Once I connected the Ultrasonic sensors to the board as well, the board would not power and the voltage regulator got extremely hot. I would assume the total current draw was too high from the outputs and VCC of the Arduino. </p>

<p>I think that if I take my 9V battery and run it through a voltage regulator 7805 then I can run the VCC power of the relays with the battery instead of the VCC from the Arduino. This would result in me only running a VCC to the 4 ultrasonic sensors, the triggers for the ultrasonics from the Arduino, and the output signals fora maximum of 2 channels per relay board resulting in 4 relays. Most of the time it is only 2 relays. </p>

<p>Would I be able to power the relay boards VCC with the 9V running through a 5v regulator instead of the arduino so that I can reduce the current running through the Arduino? </p>
 313: <p>It sounds like a power-to-ground short circuit, either a fault in the Ultrasonic sensor or in the connection to the Arduino.</p>

<p>Check your connections, and if you have an Ohm-meter, check the resistance between the sensor's power and ground pins.</p>

<p>Current consumption is a function of the device being powered, not of the power-supply, unless the supply lacks the capacity to provide as much current as the device needs.</p>

<p>Note that 9v batteries (the small brick aka a "transistor battery") is not well suited to supplying much current and probably will not power your Mega, or not for long, anyway. They are much better suited for long-term powering of very low current devices, such as a smoke detector.</p>
 42474: <p>What could be going on here apart from what @JRobert has suggested, is, LM7805 could be generating a lot of heat loss.</p>

<p>4 X Ultrasonic sensors (SR HC-04) would draw 60mA.</p>

<p>2 X Relays (SRD-05VDC-SL-C) would draw 71.4 X 2 i.e 142.8mA.(Depends on from where you signal it)
The Arduino mega would draw anywhere between 200 - 500 mA.(even more)</p>

<p>The amount loss you will have is (9-5)V * (202.8 + 250)mA 
i.e. around <strong>2 Watts</strong>. <strong>This is a lot of heat</strong>. I'm not even sure the LM7805 would be able to handle it.</p>

<p><strong>Solution</strong>: use a switching regulator like the one mentioned here: <a href="https://electronics.stackexchange.com/questions/21131/efficient-low-power-regulation-i-e-9-5-volts">https://electronics.stackexchange.com/questions/21131/efficient-low-power-regulation-i-e-9-5-volts</a></p>
