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Saturday, November 3, 2018

The diagram shows part of a current-carrying circuit. The ammeter has negligible resistance.


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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