NCEA LEVEL 1 PHYSICS
The circuit shown has four identical ammeters. A2reads 0.2A and A3reads 0.3A
What would the current be through the other two ammeters?Click here for answer
They would both read the same value - and that would be 0.5A.
The sum of the 2 currents in the parallel pathways is equal to the current before it split (0.5A). When the pathways rejoin the current is again 0.5A. Current is NOT used up around a circuit.
Which one of the following statements is correct?
- The resistance of P is more than 20 ohms.
- The resistance of P is equal to 20 ohms.
- The resistance of P is less than 20 ohms.
Give a reason for your choice.Click here for answer
The resistance of P is more than 20 ohms.
Since less current flows through A2 than A3, P has more resistance than 20 ohms.
(If they had been the same resistance then the current through each ammeter would have been the same. If P had had a lower resistance than 20 ohms then more current would have flowed through that pathway).
Write down the equation that links current, potential difference and resistance.Click here for answer
V=IR where V is voltage or potential difference, I is current and R is resistance.
Calculate the reading on the voltmeter. Show clearly how you work out your answer.Click here for answer
I=0.3A, R = 20 ohms
Since V=IR then V=0.3 x 20 = 6V
State the potential difference of the power supply.Click here for answer
This would also be 6V.
Calculate the resistance of resistor P. Show your working.Click here for answer
The voltage across P would also be 6V. The current through P is 0.2A.
V=IR so R=V/I. R=6/0.2 = 30 ohms.
(Look back at Q2. yes we said the resistance of P was > 20 ohms!)