Calculating the current through this load resistor

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SUMMARY

The discussion focuses on calculating the current through a load resistor in a circuit involving a voltage source and internal resistance. The user initially overcomplicates the setup but is guided to simplify it by placing the voltage source in series with its internal resistance. The equivalent resistance is calculated as 4,500,000 ohms, leading to a current calculation using Ohm's Law, resulting in 0.0004 A or 0.4 mA.

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Homework Statement
Find current flowing in load resistor
Relevant Equations
V=IR
Hi everyone

950347CA-A9A0-46A8-BAE8-F5B65BB3DDC8.jpeg

I have drawn what I think what the setup should look like, and worked out the equivalent resistance of the parallel connected resistors

15E10EFB-91AA-455F-A6AA-D8371C3D433F.jpeg


Not so sure where to go next with this problem?

Any help would be greatly appreciated!
 
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I think you might have overcomplicated the setup slightly. The voltage source can be drawn in series with its internal resistance, and the two ends of this arrangement connect to the load resistor. Then it should be straightforward to calculate the current using Ohm's law.
 
etotheipi said:
I think you might have overcomplicated the setup slightly. The voltage source can be drawn in series with its internal resistance, and the two ends of this arrangement connect to the load resistor. Then it should be straightforward to calculate the current using Ohm's law.

If so then then I have 2 resistors in series in the circuit. So the total resistance becomes 500,000 ohms + 4,000,000 ohms = 4,500,000 ohms

Giving I = V/R = 1800/4,500,000 = 0.0004 A or 0.4 mA?
 

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