Q#32 (Past Exam Paper – June 2016 Paper 11 Q35)
There is a current from P to R in the resistor network shown.
The potential difference (p.d.) between P and Q is 3 V.
The p.d. between Q and R is 6 V.
The p.d. between P and S is 5 V.
Which row in the table is correct?
Solution:
Answer: A.
Current flows from P to R.
Recall that current cannot flow in opposite directions in the same wire.
There are different paths that the current could possibly flow.
Path 1: P to Q to R
Path 2: P to Q to S to R
Path 3: P to S to R
Path 4: P to S to Q to R
The p.d. between P and R is constant. This means that the sum of p.d. in any of the paths should be the same.
Let the p.d. between Q and S be QS and the p.d. between S and R be SR.
Considering the different paths,
Path 1: p.d between P and R = 3 + 6 = 9 V
Path 2: p.d. between P and R = 3 + QS + SR
The sum of p.d. in path 2 should also be 9 V as the p.d. between P and R is constant.
3 + QS + SR = 9
QS + SR = 9 – 3 = 6
QS + SR = 6 eq (1)
Path 3: p.d. between P and R = 5 + SR = 9
SR = 9 – 5 = 4 V
Substituting SR = 4 V in eq (1),
QS + 4 = 6
QS = 6 – 4 = 2 V
Path 4 may not be a possible path as the sum of p.d. is different.
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