What is the current in the resistor?

You just figure it by current through each branch circuit for the information you are giving

Current through R1 branch is:

Volts / ohms = current: ohm's law:

246V / 15 ohms = 16.4 amps:

current through branch with 24 ohm resistor in series with two parallel 15 ohm resistors will be:

Total resistance of the two parallel resistors using sum over product method is>

(resistor 1 * resistor 2) / (resistor 1 + resistor 2)

(15 * 15) / (15 +15) = 225 / 30 = 7.5 ohms.

since they are in series with the 24 ohm resistor you add them together:

24 ohms + 7.5 ohms= 31.5 ohms:

Current through that branch is going to be:

Volts / ohms = amps:

246 / 31.5 = 7.81 amps.

adding the current through branches R1 and R2 gives you:

7.81 + 16.4 = 24.21 amps:

To find the resistance sub total through those two branches

Volts / current = resistance ohm's law:

246 / 24.21 = 10.16 ohms through those two branches:

What that means is you are going to have resistance in series in the circuit such as internal resistance of the power source: i.e. battery or generator:

If not that then you will have a voltage dropping resistor between point A and the branch circuits or point B and the branch circuits:

If such is the case then after finding the total resistance between point A and point B and substracting it from the sub-total resistance of branch cirucits R1 and R2, You will have to calculate the voltage across the snubbing resistor before figuring current through each branch.

Now if that is the case to find the resistance of the snubbing resistor subtract tolal resistance from sub total branch resistance:

First find total resistance:

246 / 24.2307 = 10.15 amps. total current:

combined resistance of branches R1 & Rs is: 10.16ohms

Now subtract that from total resistance:

24.2307 - 10.16 = 14.0707 for the series resistance {snubbing resistance}

again apply ohms law: V = I^2 * R to get the voltage lost across the snubbing resistor:

V= 10.15 x 14.0707=142.82V across snubbing resistor:

so voltage at branches R1 and R2 will be:

246 - 142.82 = 103.18 V

Now use branch resistance we have already found to calculate current through each branch:

103.18 V / 15 ohms = 6.88 amps through R1:

103.18 V / 31.5 ohms = 3.28 amps through R2 branch:

Just to check for errors add two branch currents together;

6.88 + 3.28 = 10.16 amps which is only .01 off due to rounding.

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