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DC circuit June 08 As

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6.(a) P=V^2/R
R=V^2/P
R=240^2/(1.5*10^3)
R=38.4 ohm

6.(b) When only S1 is open, circuit is not complete so no power.
When both S1 and S2 are closed, current passes through A only. So total power = power of A = 1.5
When all 3 are closed, current passes through A and B only. Calculate the resistance of the circuit = Resistance in parallel between A and C = (1/38.4 + 1/38.4)^-1 and again use P=V^2/R. It will give 3. Or simply since current passes through A and C, power will be the sum of the 2.
When S1 only is closed, current passes through A and B. So total resistance = 76.8. So power = 240^2/76.8 = 0.750 kW.
When S1 and S3 are closed, current passes through all 3-A,B and C. Through A and B we have found in the previous one i.e. 0.750. Now add the power of B since it is in parallel to both. Or calculate the effective resistance. Either way you'll get 2.25.

Hope it helped.
 
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Thank i understand most of it except, when S1 and S2 closed, why the current does not flow through B? Isn't B still connected?
 
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link318 said:
Thank i understand most of it except, when S1 and S2 closed, why the current does not flow through B? Isn't B still connected?


B is still connected but current have another alternative route i_e through S2 where there is no resistance to current flow so it will pass through s2 rather than resister B.......

Hope you get that!!!
 
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thanks it help. Does that apply to other conductors like, thermistor, bulb?

In a normal circuit, is it the same, meaning if there is another branch, there wont be current flowing through it?
 
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