How to calculate the current of a circuit if a resistor is placed with load in series?

Thread Starter

mannok

Joined Apr 5, 2020
9
Assume we provide a load with its rated voltage, suppose it should draw the current it need at such voltage level. However, what if we place an additional resistor in series with the load, can the load still draw the same current as it needed?
 

djsfantasi

Joined Apr 11, 2010
6,362
In a series circuit, current remains constant through the series circuit. The voltage drop is what changes across the load and resistor. So, while your load may receive the same current, it will not receive the same voltage.

if I is the current drawn by the load, R is the resistance of the resistor and Vs is the supply voltage, the resistor will drop IxR and your load will receive (Vs-(IxR)) volts.
 

Thread Starter

mannok

Joined Apr 5, 2020
9
In a series circuit, current remains constant through the series circuit. The voltage drop is what changes across the load and resistor. So, while your load may receive the same current, it will not receive the same voltage.

if I is the current drawn by the load, R is the resistance of the resistor and Vs is the supply voltage, the resistor will drop IxR and your load will receive (Vs-(IxR)) volts.
Hey thank you @djsfantasi. That means the current would not change while the load receive less voltage right?

Base on the above circumstance, Assume I have a power source which is capable to output a large number of voltage (e.g. 1000V) and it can limit the output current to how much the load needs. Does it mean that the resistor won't affect either the voltage or the current taken by the load? Since the load "draws" current it needed at certain voltage. In this case, the load can draws as much current as the power source can supply, at the same time, the power source can provide "more than enough" voltage to the circuit base on the current drew by the load.
 

dl324

Joined Mar 30, 2015
10,536
Welcome to AAC!
if we place an additional resistor in series with the load, can the load still draw the same current as it needed?
If the load was a current source sink and the power supply provided sufficient voltage for it to function, the current would stay the same. If the load was a resistance, the current would decrease.

It would be helpful if you posted a schematic for the situation so we can tell what the load is and give you a more specific answer.
 

ericgibbs

Joined Jan 29, 2010
9,853
hi mannok,
If the load is connected to a fixed voltage source and you add a series resistor with the load , the current thru the load will be reduced.
E
 

Thread Starter

mannok

Joined Apr 5, 2020
9
Welcome to AAC!
If the load was a current source sink and the power supply provided sufficient voltage for it to function, the current would stay the same. If the load was a resistance, the current would decrease.

It would be helpful if you posted a schematic for the situation so we can tell what the load is and give you a more specific answer.
Thank you @dl324 , thank you for you help... How do I distinguish if it is a current sink or a resistance? Take motor as an example, it can be a current sink and resistance at the same time.

btw... yesyes, I am going to post the schematic...
 

ElectricSpidey

Joined Dec 2, 2017
1,070
Assume we provide a load with its rated voltage, suppose it should draw the current it need at such voltage level. However, what if we place an additional resistor in series with the load, can the load still draw the same current as it needed?
Only if you raise the supply voltage.
 

Thread Starter

mannok

Joined Apr 5, 2020
9
Thank you @dl324 , thank you for you help... How do I distinguish if it is a current sink or a resistance? Take motor as an example, it can be a current sink and resistance at the same time.

btw... yesyes, I am going to post the schematic...
circuit.jpeg

The background of my question can refer to this schematic... I have a load in the collector of a BJT, the voltage it takes is depend on the current it draws by the amplified version of base current. Therefore, I am just thinking of "does it means that the value of the resistor in the collector won't affect how the load operate?"
 
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dl324

Joined Mar 30, 2015
10,536
I have a load in the collector of a BJT, the voltage it takes is depend on the current it draws by the amplified version of base current.
That circuit won't work well. You seem to be trying to convey that the load wants a 10V supply, but you're connecting it to a 1000V supply. You want to use a transistor to control power to the load, but the transistor can't be used as a switch. If you want the collector-emitter voltage to be 990V at 1A (leaving 10V for the load and ignoring the resistor which doesn't have a value), the transistor would need to dissipate 990W.
 

Thread Starter

mannok

Joined Apr 5, 2020
9
That circuit won't work well. You seem to be trying to convey that the load wants a 10V supply, but you're connecting it to a 1000V supply. You want to use a transistor to control power to the load, but the transistor can't be used as a switch. If you want the collector-emitter voltage to be 990V at 1A (leaving 10V for the load and ignoring the resistor which doesn't have a value), the transistor would need to dissipate 990W.
Yes I know it doesn't make sense in reality. However, In this circuit I just want to get my self clear about why some loads can control/draw a constant value of current from power supply while some cannot (a resistor is indeed a kind of load, but current drawn is depending on L but not R1). What makes a load dominate the control of current? Also, what will happen if 2 loads are run in series and they have different rating value of current at same voltage? Which one will dominate the control of current in such case?
 

dl324

Joined Mar 30, 2015
10,536
However, In this circuit I just want to get my self clear about why some loads can control/draw a constant value of current from power supply while some cannot (a resistor is indeed a kind of load). What makes a load dominate the control of current?
You're going to need to give some specific examples.

Most circuits are designed for a specific range of input voltages. Going outside of that range shouldn't even be considered unless you think you know more than the person who designed it.
 

Thread Starter

mannok

Joined Apr 5, 2020
9
You're going to need to give some specific examples.

Most circuits are designed for a specific range of input voltages. Going outside of that range shouldn't even be considered unless you think you know more than the person who designed it.
Sure let me make it simple...
In such case, what will be the current in the circuit?

Seems that motor has the ability to control the value of current. When there are 2 motors then which one "take the control"?
 

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LesJones

Joined Jan 8, 2017
2,606
Also a motor does note take a fixed current. The current changes in line with the mechanical load on the motor. Tell us what you are trying to do. The only situation I can think of for connecting 2 motors in series would be using two IDENTICAL motors when you wanted the motors to provide equal torque. This would be true for permanent magnet motors or shunt wound motors with the armatures connected in series but the fields fed from a different supply.

Les.
 

BobTPH

Joined Jun 5, 2013
2,351
You need to know the IV curve of each load in series, then it comes down to a system of simultaneous equations that you can solve.

In your case of motors, there is a an added complication of the mechanical load. If that remains constant, the two motors you have shown will split the voltage equally. If you load down one of the motors (think stopping it from rotating,) then the current in both motors will go up and the voltage will go down on the stopped motor and up on the one running freely, perhaps enough to destroy it.

Bob
 

MrAl

Joined Jun 17, 2014
7,511
Assume we provide a load with its rated voltage, suppose it should draw the current it need at such voltage level. However, what if we place an additional resistor in series with the load, can the load still draw the same current as it needed?
Not if it is connected to a constant voltage source.
 
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