Question

Three resistor 3,4 and 5 ohms are joined in a parallel in an electric circuit.If a current of 150mA flow through resistor 4 ohms then calculate the value in mA which will be flowing through other two resistors.

Answer 1

The battery's current will be exclusive to the battery’s specs and can by no means be predicted using available data such as this for calculation.

Circuit current will be the 1-amp cited by others herein.

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Answer 2

R1 = 4ohm
R2 = 6ohm
R3 = 12ohm
Rs = 2ohm (Series resistor)
Potential difference,V = 6V
Let the resistance of all three resistors which are connected in parallel be Rp
Then,
1/Rp = 1/R1 +1/R2 + 1/R3
1/Rp = 1/4 + 1/6 + 1/12
1/Rp = 6/12
or,Rp = 12/6 = 2ohm
Let total resistance be Rt = Rp +Rs
Rt = 2 + 2 = 4ohm
a).So current,I in the main circuit will be V=IR
therefore, I = V/R = 6/4 = 1.5A which is total current.
b) Since in parallel the current will divided into three resistors connected in parallel.Let the current across all the three be I1,I2 and I3.
Now,
I1 = V/R1 (Since potential difference is same as they are connected in parallel)
I1 = 6/4 = 1.5A
Similarly,I2 = V/R2
I2 = 6/6 = 1A
and I3 = 6/12 = 0.5A
c). V across 2ohm resistor will be-
V = I*R ( I will be total current)
V = 1.5 * 2 = 3V