Resistance - voltage drops please help?

Question

I have 5 volts battery at the bottom and three resistors - 2 in series parallel to the battery (1 and 2 ohms) and one above them parallel to them (3 ohms). Please help me to calculate voltage drop accross each one.
How to I calculate current through those two lines - I know the total current and I know that the sum of those two is total. Helppppp!

Answer 1

Is this the circuit?

---------3 ohm------------
l *********************** l
l *********************** l
------ 1 ohm----2 ohm--
l *********************** l
l *********************** l
----------5 Volts-----------


If yes,
here's how to solve

First, the voltage across the 3 ohm resistor is 5 volts since it is in parallel to the battery

the series resistors sum, to 3 ohms

since the voltage across the pair is 5 volts,
and V=I*R
I=5/3

This s also the current through the 3 ohm resistor in parallel.

the voltage across the 2 ohm
is V2=2*5/3
and the one is
V1 = 5/3

If I had the diagram incorrect add comments and I'll correct.

BTW: I'm 49

j

Answer 2

Look, this is not a very difficult exercise, but, you really need to make a graphic, and you will have, two resistors in series and one above in parallel with those two resistos, and above the 3 resistors, you have the battery.

Hint one : Find the equivalent resistor :

Two in series : Req = 1 + 2 = 3

and working with the parallel one : 1 / 3 + 1 / 3 = 1 / Rt

Rt = 3 / 2 (ohms)

Now, we can find the total current, with the battery, let's apply the Ohms ' law :

5 = I*3/2

I = 10 / 3 (Amperes)

When you have resistors in series the current is the same, when you have resistors in parallel, the potential difference is the same, in this case 5 volts.

5 volts, is the potential difference between the 3 oms resistor and between the 2 resistors in series ( 1 ohm and 2 ohm )

Using 5 volts and the resistor of 3 ohm, we can find it's current :

5 = 3*I'

I' = 5/3 amperes >>> current in the 3 ohm resistor.

To find the current on the other resistors ( series resistors ), just do this :

total current - 5/3, why ?

because, the current is moving through those two lines, so the current must be the same, right ?

so : 10 / 3 - 5/3 = 5/3 amperes

On each line, you will find 5 / 3 amperes, and the total current will be 10 / 3 Amperes

Answer 3

Inthis case the series resistors sum up the same as the paralel resistor so the equivalente resistence is 1.5 ohm and the total current5/1.5 amp which flows equally divided through both lines
So the voltage drop through the 3 ohm resister is 5/(2*1.5) *3 = 5V as should be
The voltage drop through 1ohm is 5/3 *1 V
and 5/3 *2 V though the other

Answer 4

look, it fairly isn't an fairly complicated exercising consultation, yet, you fairly favor to make a photo, and also you're turning out to be to be to be, 2 resistors in sequence and one above in parallel with those 2 resistos, and above the three resistors, you've the battery. hint one : hit upon the equivalent resistor : 2 in sequence : Req = a million + 2 = 3 and dealing with the parallel one : a million / 3 + a million / 3 = a million / Rt Rt = 3 / 2 (ohms) Now, we are able to hit upon the large cutting-part-day, with the battery, enable's practice the Ohms ' regulation : 5 = I*3/2 I = 10 / 3 (Amperes) upon getting resistors in sequence the present is an same, upon getting resistors in parallel, the potential massive distinction is an same, subsequently 5 volts. 5 volts, is the potential massive distinction between the three oms resistor and between both resistors in sequence ( a million ohm and a couple of ohm ) utilizing 5 volts and the resistor of three ohm, we are able to hit upon it fairly is cutting-part-day : 5 = 3*I' I' = 5/3 amperes >>> cutting-part-day interior the three ohm resistor. to discover the present on the diverse resistors ( sequence resistors ), do precisely this : finished cutting-part-day - 5/3, why ? as end results of the technique of truth, the present is transferring by technique of creating use of those 2 strains, so the present may be an same, ideas-blowing ? so : 10 / 3 - 5/3 = 5/3 amperes On each and each and each and each line, you'll discover 5 / 3 amperes, and the large cutting-part-day could be 10 / 3 Amperes