A battery is connected to a 2.3Ω and a 4.2Ω resistor connected in parallel.?

Question

The current through the battery is 2.0 A. Find the current in the 2.3Ω resistor?
Can someone help me tackle this problem?

Answer 1

Use the current division method:

For two(2) resistor that are connected in parallel, say R1 and R2, and given source current, say I and current that past through R1 and R2, say I1 and I2 respectively, the formula would easily be :

I1 = I(R2 / (R1 + R2) )

I2 = I(R1 / (R1 + R2) )

or simply...

= I - I1

So, as for your question :
Let : I = 2.0 A
R1 = 2.3 ohm
R2 = 4.2 ohm

thus, current flow in the 2.3 ohm resistor = I1

I1 = I(R2 / (R1 + R2) )
= 2.0 (4.2 / (2.3 + 4.2) )
= 1.29 A ... ans

Answer 2

Use the current division method:

For two(2) resistor that are connected in parallel, say R1 and R2, and given source current, say I and current that past through R1 and R2, say I1 and I2 respectively, the formula would easily be :

I1 = I(R2 / (R1 + R2) )

I2 = I(R1 / (R1 + R2) )

or simply...

= I - I1

So, as for your question :
Let : I = 2.0 A
R1 = 2.3 ohm
R2 = 4.2 ohm

thus, current flow in the 2.3 ohm resistor = I1

I1 = I(R2 / (R1 + R2) )
= 2.0 (4.2 / (2.3 + 4.2) )
= 1.29 A ...

Answer 3

For PARALLEL resistors the voltage drop around the resistors is the same for the two. think of roughly it. you may no longer have 2 distinctive voltages interior the same cord. You added an added course so #a million is erroneous no longer necessarly so #2 is erroneous no longer necessarly so #3 is erroneous Yep, #4 is right

Answer 4

Call the battery voltage Vb. The current through the 2.3-ohm resistor is Vb/2.3. The total current is Vb/2.3 + Vb/4.2 = 2A; solve for Vb and plug it into the first eq.

Answer 5

I'm not sure.. since i dropped out of Physics 208.. but i think the voltage stays the same in a parallel circuit.. so
V=IR
plug in your known 2.0A and 6.5ohm as R
V=(2)*(6.5)=13V
Then solve for I=V/R
I=(13/2.3)=5.65A