1) How much electricity flows into the network of bulbs A, C and D? 2) How much
ID: 1537281 • Letter: 1
Question
1) How much electricity flows into the network of bulbs A, C and D?
2) How much of the flow passes through bulb C?
**** 3) What can you say about the flow through bulb A? ( give your answer in terms of f ) and Explain << ** Need help with this one the most! please work out how you came to the final equation for the flow of bulb A. In final format.. f = ________
Recall that we assumed that the flow out of the network of bulbs B and E is f. To help with the network of bulbs A, C and D, it might help if you think about the following, Circuit 1 D Only the network of A, C & D has changed. Actually, bulbs C and D are still there. What is new is that bulb A has been replaced with two pairs of(identical) bulbs in parallel. Begin by convincing yourself that two pairs of identical) bulbs in parallel with each other presents the same obstacle as a single bulb. The reason for doing this is that now all three branches are the same as the branch that includes C and D. You already know how much electricity flows into the network of bulbs A, C and D. Recalling the assumption that is underlined above,Explanation / Answer
let resistance of each bulb be R.
let voltage supply be V.
then for the left side circuit:
there are two bulbs in series in each branch.
net resistance=R+R=2*R
there are 3 such branches in parallel.
net resistance=2*R/3
B and E in parallel present a resistance of R*R/(R+R)=R/2
so total resistance in the circuit=(2*R/3)+(R/2)
=(7/6)*R
current through the circuit=voltage / total resistance
=6*V/(7*R)
this is current that is flowing out of B and E.
hence f=6*V/(7*R)
now due to current division , flow through bulb A (which has a resistance of R and is in parallel with 2*R ie. series combination of C and D)
=f*(2*R)/(2*R+R)
=2*f/3
hence flow through bulb A is 2*f/3.
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