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NTU Electrodynamics

Questions and solution for  nanyang technological University Singapore Entrance Examination preparation in physics on the syllabus of Electricity and magnetism - Current of electricity.

Question 1
Given an electric circuit below



What is the magnitude of current in the 40 Ω resistor?
A. 0.41 A
B. 0.33 A
C. 0,77 A
D. 1.30 A


Solution
We may use the shortcut to solve this question.



The first find the total resistance  for the circuit:




Then the potential difference between a and b points:



Applying kirchoff law to find I

Question 2

Given an electric circuit shown below fugure.



In the circuit, the power dissipated in the 6 Ω resistor as heat is 6 watt. What is the value of resistance X in above circuit?

Solution
The power P which is dissipated in 6 Ω is 6 watt.




P = I2R
6 = I2 (6)
I = 1 Ampere

The voltage across 6 Ω or VAB = IR = (1A)(6 Ω) 6 volt. So then the voltage across BC, VBC = 12 volt − 6 volt = 6 volt.

X and 8 Ω in parallel connection so then the voltage across is the two resitors are same value and  the currents pass:
I2 = V/R = 6 / 8 = 3 / 4 = 0.75 A
I1 = I − I2 = 1 − 0,75 = 0.25 A

The resistor X = V/I = 6 / 0.25 = 24.0 Ω

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