(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
<h3>What is current?</h3>
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.
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