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Electricity question Watch

1. A has a value of 2 ohms B a value of 4 ohms, the cell is 2 volts and 1 ohm

I need to calculate the current in A and B and the p.d. between the terminals of the cell.

For the Current in A i did I=v/R for the cell so 2/1 = 2 and we know ina parallel the voltage stays the same so its going to be 2/2 = 1 amp, but in the answers they get 0.47A, similalry with B

C im not too sure about
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2. Hello Hazbaz

in your fig, and taking C to be 2 ohm, we have a circuit something like this.

(joints the other end/earth)------[cell]------[A]------[B // C]------(joints the previous end/earth)

= ()------[Vcell ideal+ Cell Resistance]------[A]------[B // C]------()

Now, total resistance [TR] of the system = [CR] + [A] +[B//C] = [CR] + [A] +{B*C/(B+C)}

= 1+2+[4//2] = 3 + (8/6) = 26/6

cell potential / total resistance gives the total current or the current flowing through [A]. = 0.46 A

from this you should be able to calculate current through B.

potential across B = potential across C

Ib*Rb = Ic*Rc

Ib +Ic = 0.46

You should be able to work out this to find Ib

Now the potential across [TR] =ideal cell voltage = 2Volts.

[TR] = cell resistance (1ohm) + external effective resistance (20/6)ohms ([A]+[B // C])

of which the potential across 1 ohm is lost within the cell, and the potential across {[TR]-1} is avaliable cell voltage.(potential across the cell)
or simply, a current of 0.46A through 1ohm resistance drops/develops 0.46V. The remaining is the avaliable cell voltage.

I hope now you can work out own your own.

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Updated: December 8, 2010
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