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Six Resistors Are Connected As Shown In Fig. 2.24. If A Battery Having An E.M.F. Of 24 Volts And Internal Resistance Of 1 S2 Is Connected To The Terminals A And B, Find (I) The Current From The Battery, (Ii) P.D. Across 8 12 And 4 12 Resistors And (Iii) The Current Taken From The Battery If A Conductor Of Negligible Resistance Is Connected In Parallel With 8

Respuesta :

(I) The Current From The Battery is 3A.

(Ii) P.D. across 8 ohm and p.d across 4 ohm is 12V and 3V respectively.

(Iii) The Current Taken From The Battery If A Conductor Of Negligible Resistance Is Connected In Parallel With 8Ω is 6A

What is current?

The current is the stream of charges which flow inside the conductors when connected across the end of voltage.

From the Ohm's law, V =IR

R = V/I

Here, R is the proportionality constant.

i)

The resistance 4Ω and 2Ω are in series, So their equivalent is 4+2 =6Ω.

The 6Ω and 6Ω are in parallel, so  their equivalent is 6x6/6+6 = 3Ω.

The  3Ω and 5Ω are in series, so their equivalent is 3 +5 = 8Ω.

The 8Ω and 8Ω are in parallel, so their equivalent is 8x8/8+8 = 4Ω.

The 4Ω and 3Ω are in series, so their equivalent is 4 +3 = 7Ω

The 7Ω and 1Ω are in series,  so their equivalent is 7+1 =8Ω

So, the total equivalent resistance is 8Ω.

The current from the battery is  I =V/Req

I  = 24/8 = 3A

Thus, the current from the battery is 3A.

II) Applying the mesh analysis in three loops, we have

-24 +1 i1 +3i1 +8(i1-i2) =0

12 i1 -8i2 =24..................(1)

5 i2 +6(i2 - i3) +8(i2 -i1) = 0

-8 i1 +19 i2 -6 i3 =0........................(2)

2 i3 +4 i3 +6( i3 - i2) =0

-6 i2 +12 i3 =0...................(3)

On solving these equations, we get

i1 = 3A

i2 = 1.5A

i3 = 0.75A

p.d across 8 ohm = V8 =(i1 - i2) x 8 = 12V

p.d across 4 ohm = V4 = i3 x 4 = 3V

Thus, the p.d across 8 ohm and p.d across 4 ohm is 12V and 3V respectively.

iii)The resistance 2 and 4 ohm are in series, them equivalent is 2 +4 =- 6 ohm

The 6 and 6 ohm are in parallel, so their equivalent is 6x6/6+6 = 3 ohm

The 3 and 5 ohm are in series, so their equivalent is 3 +5 = 8 ohm.

Now 8 ohm is in parallel with a short circuit path. Whenever a short circuit  appears , current will flow through it. So resistance  cant be taken into consideration.

Then, Req = 3+1 ohm = 4 ohm

The current flow will be

I = 24/4 = 6 A.

Thus, The Current Taken From The Battery If A Conductor Of Negligible Resistance Is Connected In Parallel With 8 ohm is 6A.

Learn more about current.

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