A potentiometer circuit is used to determine the electromotive force (e.m.f.) $$\(E\)$$ of a cell. The circuit includes a second cell of e.m.f. 1.5 V and internal resistance $$\(0.50 \Omega\)$$ that is connected to a uniform resistance wire XY, as shown. The resistance wire XY has a length of 0.96 m and a resistance of $$\(0.50 \Omega\)$$. The movable connection $$\(Z\)$$ is moved along wire $$\(X Y\)$$. The galvanometer reading is zero when length XZ is 0.64 m . What is the value of e.m.f. $$\(E\)$$ ?
A.
0.50 V
B.
0.75 V
C.
1.0 V
D.
1.1 V
Exam No：9702_w24_qp_11 Year：2024 Question No：37
Answer:
A
Knowledge points:
10.2.1 recall Kirchhoff’s first law and appreciate the link to conservation of charge
10.2.2 recall Kirchhoff’s second law and appreciate the link to conservation of energy
10.2.3 derive, using Kirchhoff’s laws, a formula for the combined resistance of two or more resistors in series
10.2.4 solve problems using the formula for the combined resistance of two or more resistors in series
10.2.5 derive, using Kirchhoff’s laws, a formula for the combined resistance of two or more resistors in parallel
10.2.6 solve problems using the formula for the combined resistance of two or more resistors in parallel
10.2.7 apply Kirchhoff’s laws to solve simple circuit problems
Solution:
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