If the battery voltage is 120v and the resistances are...

Name: if the battery voltage is 120v and the resistances are r1100omega r2600omega and r3130omega a what is the total resistance b find the current c calculate the voltage drop in each resistor an 28012
Uploaded: 2020-07-02T07:06:57-08:00
Duration: 6 min 57 s
Channel: Ren Jie Tuieng
Description: if the battery voltage is 120v and the resistances are r1100omega r2600omega and r3130omega a what is the total resistance b find the current c calculate the voltage drop in each resistor an 28012

If the battery voltage is 12.0V, and the resistances are R1=1.00Omega ,R2=6.00Omega and R3=13.0Omega (a) What is the total resistance? (b) Find the current. (c) Calculate the voltage drop in each resistor and show these add to equal the voltage output of the source. (d) Calculate the power dissipated by each resistor. (e) Find the power output of the source, and show that it equals the total power dissipated by the resistors 5.If the battery voltage is 12.0V,and the resistances are R1=1.00O,R2=6.00O and R3=13.0O N R3 (a) What is the total resistance? (b) Find the current. (c) Calculate the voltage drop in each resistor and show these add to equal the voltage output of the source (d) Calculate the power dissipated by each resistor. (e) Find the power output of the source, and show that it equals the total power dissipated by the resistors

Transcript

00:01 Now in this question we're interested in this particular circuit and we have quite a few things to find about this circuit.

00:08 The first one would be the equivalent resistance of the entire circuit.

00:14 To do that, we have to simplify this parallel circuit over here first by taking 1 over r2 plus 1 over r3 and we inverse it.

00:29 So that's 1 divided by 9 plus 1 divided by 18.

00:33 Get 6 oms then now the others all in series with each other so we can just add up the resistance so the effective resistance which is 6 plus 14 plus 4 oms to give us 24 oms so there's the effective resistance now to find the current we got to use i equals to v over r where v is the total voltage across the entire circuit.

01:16 So we know that the batteries v1 and v2 are actually connected in the same direction.

01:23 So we can assume them to be a single large battery with total voltage of 24.

01:31 We add them up together because they are in series with each other.

01:34 So this will be 24 volts, divided by 24 oms.

01:39 To give us 1m.

01:43 This is our current is flowing out of the batteries.

01:50 Now to find current through each resistor we already have that the current flowing out the battery is 1 amp so this is 1m.

01:59 So i1 flowing through r1 must be 1 amp and i4 flowing through r4 must also be 1m right since this is the current of the batteries.

02:12 Now what about the current for our r2 and r3? well, one of the ways that we're going to do is we can find what is the potential difference across this r2 and r3 first by using the resistance, which is equal to 6 oms.

02:33 You could learn resistance of 6 oms.

02:36 So the potential difference across r2 and r3, just called v prime, will be equals to f equals to ir so we take the total current which is 1 amp multiplied by 6 and we get 6 volts next from here we can find the individual current so i2 must be equals to 6 volts developed by 9 should be 0 .67 amps and from here we can infer that our i3 will be just 1 minus this value.

03:35 Alright, so now we have the current of each resistor...

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