Question 434: [Current of Electricity > Potential difference] (Past Exam Paper – June 2007 Paper 1 Q33)
Cell of e.m.f. 2.0 V and negligible internal resistance is connected to network of resistors shown.
V1 is potential difference between S and P. V2 is potential difference between S and Q.
What is the value of V1 – V2?
A +0.50 V B +0.20 V C –0.20 V D –0.50 V
Solution 434:
S is connected to the negative terminal of the cell, so it is at a potential of 0V.
From Kirchhoff’s law, the sum of p.d. in each loop is equal to the e.m.f. of the battery.
Consider the loop containing point P.
Since the 2 resistors have the same resistance, the p.d. across each of them will be equal.
So, the potential difference between point P and S is 1.0V and since point S is at a potential of 0V, point P should be at a potential of 1.0V. V1 = 1.0V.
Consider the loop containing point Q.
From the potential divider equation, the p.d. across the 3.0kΩ resistor is
p.d. V2 = [3 / (2+3)] x 2 = 6 / 5 = 1.2V
Thus, V1 – V2 = 1.0 – 1.2 = -0.2V
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