Question 16
The diagram shows part
of a current-carrying circuit. The ammeter has negligible resistance.
What is the reading on
the ammeter?
A 0.7 A B 1.3
A C 1.5
A D 1.7
A
Reference: Past Exam Paper – June 2015 Paper 13 Q37
Solution:
Answer:
C.
Let the total resistance of the circuit be R.
1/R = 1/1 + 1/2 + 1/5
giving
Total resistance R = 0.588 Ω
In other words, the three resistors in parallel is
equivalent to a single resistor of resistance 0.588 Ω.
A current of 5.0 A flows through this resistor.
p.d. across the branches = IR = 5.0 × 0.588 =
2.94 V
As the resistors are in parallel, the p.d. across each
branch is the same (= 2.94 V).
V = IR giving
I = V/R
Current through ammeter = V / R = 2.94 / 2.0 =
1.47 A = 1.5 A
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